A ball thrown vertically upward is caught by the thrower after 8.00 s.?

Question

A ball thrown vertically upward is caught by the thrower after 8.00 s.

(a) Find the initial velocity of the ball.
m/s
(b) Find the maximum height it reaches.
m

Answer 1

Let time taken for ball to reach highest point be t
Initial velocity = u
Final velocity (v) = 0 (Ball stops after it reaches highest point)
acceleration due to gravity (g) = 10 m/s^2

v = u - gt (- is used as acceleration is opposite the direction of motion)
0 = u - gt
-gt = -u
gt = u
t = u/g

Let time taken for ball to reach ground from highest point be t
It can again be proved that t = u/g

Total time taken = 8s
So, u/g = 4
u/10 = 4
u = 40 m/s
Initial velocity of the ball is 40 m/s

u = 40
height = h
g = 10
t = 4
h = ut - 1/2 gt^2
= 160 - 80
= 80 m
The ball rises to 80 m height

Answer 2

In words: considering the fact which you would be neglecting the air resistance, the flight of the ball on your occasion would be symetrical (that it, it takes as long to fall bypass into opposite because it takes to prevail in that is optimal top, and at any given top the ball reaches, that's going to be doing an identical velocity on the way down because it became into doing on the way up). Therefor, it reaches that is optimal top halfway for the time of the whole time elapsing in the past that is caught lower back (a million.35 s). you recognize that is deccelerating at 9.80 one ms^-2 jointly as that is in the air, and you recognize it takes a million.35 seconds to provide up (because of the fact on the precise of that is trajectory, that is no longer moving). So it is going to lose 9.80 one x a million.35 ms^-a million of upwards velocity in coming to a halt on the precise - as a result, it incredibly is the preliminary velocity, vertically upwards. As for the optimal top, you are able to the two think of with reference to the replace of kinetic capability and gravitational ability capability, understanding the preliminary velocity, or you are able to sparkling up equations of action. capability's many times much less complicated, so a sprint for the examination is to choose the thank you to do the capability calculations (i'm going to bypass away this to you) - because of the fact they might no longer anticipate you to, so it could make the examination questions much less complicated... utilising equations of action, you like s= uxt + a million/2 x a x t^2. 'u' is the preliminary velocity - you chanced in this above. 's' is the 'displacement', or distance the ball's long previous up. to locate 's' for the precise of the trajectory, you need to use the time this occurs at, which we labored out and used above. bear in concepts that the acceleration's damaging! Now only bang in the figures.

Answer 3

The trajectory of the ball is defined by the equation

h = v0*t - .5*g*t^2

The velocity of the ball is given by

v = v0 - g*t.

After 8 seconds the ball has gone up to max height and back down. it will reach max in half that time (up down trajectories are symmetrical).

The max height occurs when v = 0 (the ball stops rising). This happens when v0 = g*t, t is 4 seconds, solve for v0. Put this v0 and 4 seconds in the first eq to get max height.

Answer 4

(a) 0 = v - 4a
v = (4 s)(32.2 ft/s^2) = 128.8 ft/s
(*60mph/88fps = 87.8mph)
(b) h = (v^2)/2a
h = (128.8 ^2)/(2*32.2)
h = (16589.44)/(64.4)
h = 257.6 ft.

Answer 5

4.00s up, 4.00s down.
v = gt
v = (9.81 m/s^2)(4.00 s)
v = 39.2 m/s

s = 1/2gt^2
s = 1/2 (9.81 m/s^2)(4.00 s)^2
s = 78.5 m