this question about the max height above the roof?

Question

a man stands on theroof of a 15 m tall building and throw a rock with a velocity of magnitude 30m/s at the angle of 33 above the horizontal the max heigh above the roof and the velocity of the rock before it strickes the ground and the horizental distance from the base of the building to the point where the rocket just strikes the ground

Answer 1

The equations for position will be used for this solution

x(t)=x0+vx*t
y(t)=y0+vy*t-.5*g*t^2

the equations for velocity (before striking the ground) are

vx(t)=vx
vy(t)=vy-g*t

Using the data given
y0=15m
vy=sin(33)*30
x0=0
vx=cos(33)*30

the rock will reach apogee when vy(t)=0

t=sin(33)*30/9.81
=1.666 sec

the height above the roof will be

y(1.66)-15=
sin(33)*30*1.666-
.5*9.81*1.666^2

13.6 m above the roof

The rock strikes the ground when y(t)=0

0=15+sin(30)*30*t-.5*9.81*t^2
This is a quadratic with the following relevant root

t=4.08 sec

at this time the distance from where the rock was thrown is
x(4.08)=cos(33)*30*4.08
x=102.7 m

the velocity magnitude can be found two ways

Conservation of energy

.5*m*v^2=.5*m*30^2+m*g*h
where h=15
v=sqrt(30^2+2*9.81*15)
=34.6 m/s

Or evaluating
sqrt(vx(4.08)^2+yx(4.08)^2)

vx(4.08)=cos(33)*30
yx(4.08)=sin(33)*30-9.81*4.08
v=sqrt(25.16^2+(-23.68^2))

=34.6

Same answer

j

Answer 2

This question can be solved by finding the horizontal and vertical components of the rocks velocity at the start and then using the vusat equations

Draw the diagram to help you

I will use the vusat equations in my answer, with the following definitions:
v = final velocity
u = initial velocity
s = displacement during flight
a = acceleration (only gravity in this case = -9.8 m/s)
t = time

At the start, the components of the velocity are as follows:
Horizontal = 30 cos 33
Vertical = 30 sin 33

a) The maximum height reached above the roof:

the maximum height reached will be when the vertical component of the velocity is 0 (i.e. when the rock is no longer moving upwards)

v^2 = u^2 + 2*a*s

we are trying to find s ( the displacement upwards )

v^2 = 0 (as the vertical component of the velocity is 0)
and u is the vertical component of the velocity = 30 sin 33

so 0 = (30 sin 33)^2 + 2 * -9.8 * s
2 * 9.8 * s = (30 sin 33)^2
s = [ (30 sin 33)^2 ] / [ 2 * 9.8 ]

s = 13.6 m above the height at which it was thrown (the top of the building as we are ignoring the height of the man)

b) The velocity of the rock before it strikes the ground

to find this we need to find the magnitude of the resultant of the horizontal and vertical components of the velocity

so basically we find how fast it is going in each direction and use pythagoras' theorem to work out the overall velocity

without air resistance, the horizontal velocity will remain the same

to calculate the vertical velocity, we can use the same vusat equation (but are trying to find v this time, not s)
v^2 = u^2 + 2 * a * s

s = -15 metres (as this is the height change going upwards)

v^2 = (30 sin 33)^2 + 2 * -9.8 * -15
v^2 = 561
v = 23.7, -23.7 m/s
as the rock is going down at this point,
v = -23.7 m/s

so when the rock strikes the ground, the components of the velocity are:
Horizontal = 30 cos 33 = 25.2 m/s (unchanged as no air resistance)
Vertical = -23.7 m/s

so the resultant velocity can be calculated by:
sqrt[ (25.2)^2 + (-23.7)^2 ]

= 34.6 m/s

c) the horizontal distance can be calculated using
s = (u+v)/2 * t

as u and v are the same for the horizontal component of the velocity, the equation becomes
s = u * t

where t is the time taken to hit the floor, which we need to calculate

again we can use
s = (u + v)/2 * t
to find the time taken, this time with the vertical u, v and s
t = (2 * s) / (u + v)
t = (2 * -15) / (30 sin 33 + -23.7 m/s)
t = 4.08 seconds

so the horizontal distance travelled
s = u * t
s = 30 cos 33 * 4.08
s = 102.76

distance from base of building = 103 m

Answer 3

alright, so the initial velocity is 33m/s at an angle of 33 degrees. You have to separate this velocity into the vertical and horizontal components. The vertical component is the opposite (you know the hypotenuse), so Vy = 30 sin 33 = 16.339m/s. and Vx = 30 cos 33 = 25.16m/s.

Now, the horizontal velocity (Vx) is constant the whole way.
THe vertical velocity changes though because there is an acceleration that is acting downwards at a rate of 9.8m/s^2 (gravity).

To find the max height, above the roof, it is the point where the velocity is 0. The initial velocity is 16.339m/s up and it is decreasing at a rate of 9.8m/s^2. V = a*t, so t = V/a = 16.339 / 9.8 = 1.6672 seconds.

Using this time, you can find what the maximum height is above the roof. Easiest way to do this is to consider the average velocity between when he throws it and when it is 0, this is just the initial velocity divided by 2, which is 8.1695 m/s. You can use this because the velocity is changing linearly. So this average velocity multiplied by the time, will give you the distance it travels.
D = V*t = 8.1695*1.6672 = 13.62m above the roof is the maximum height.

Next part is to find the velocity of the rock before it strikes the ground.

This distance would be 13.62m + 15m. You know the initial velocity of this downward motion is 0 because the velocity at the maximum height is 0.
so h = 28.62m.

The final velocity would be the acceleration multiplied by the time, because this is equal to the change in velocity and the initial velocity is 0. So Vf = at. If you wanted to use an average velocity, you would take this velocity and divide it by 2, so V = at/2. You know that distance is equal to velocity multiplied by the time, so you have the velocity and you have to multiply by t.
so D = at^2 / 2 = 28.62m.

Solve for t to get the time it takes for the object to reach the ground.

28.62 = (1/2)*9.8*t^2
28.62/4.9 = t^2
t^2 = 5.84
t = 2.417s.

This time can be used to find the final velocity (velocity just before it strikes the ground). Final velocity is equal to acceleration multiplied by time, so 9.8m/s^2 * 2.417s
Vf = 23.68 m/s.

The total time from when the rock left the man's hand till when it hit the ground is 1.6672s (going up part) + 2.417s (going down part) = 4.0842s.

The whole time the vertical velocity was changing, the horizontal velocity was constant. So when the rock hits the ground, it has travelled at 25.16m/s horizontally for 4.0842s.

To find the horizontal distance, just multiply velocity * time.
HDistance = 102.758m horizontally.

I hope you understood all that.

Answer 4

Ok, let's start step by step :

First, let's decompose the velocity :

Vx = 30cos(33) = 25.1 m/s

Vy = 30sin(33) =16.3 m/s

The horizontal distance : Vx*T = X

X = 25.1*T

T = total flying time.

Let's work know with the vertical velocity, to find the altitude the rock reaches, and the time the rock took to go up and the to go down :

When the rock reaches its maximum altitude, it's final velocity is 0 m/s

Vf^2 = Vo^2 - 2*g*H

g = 9.8 m/s^2

0 = 16.3^2 - 2*9.8*h

h = 13.5 m

Let's find the time it took : Vf = Vo -g*t1

0 = 16.3 -9.8t1

t1 = 1.66 seconds

The max heigh above the roof : 13.5 meters

The rock is now : 28.5 meters over the ground, and with an initial velocity of 0 m/s, let's find the final velocity, just before it strikes the ground :

Vf^2 = Vo^2 + 2*g*H

Vf^2 = 2*9.8*28.5

Vf = 23.6 m/s >> Just before it strikes the ground

Let's find the time : Vf = Vo +g*t2

23.6 = 9.8 t2

t2 = 2.4 seconds

T = t1 + t2 = 2.4 + 1.66 = 4.06 seconds

X = Vx*T = 4.06*25.1 = 102 meters

Hope that helps