Calculate to what height a ball rises?

Question

A ball is shot from the ground into the air. At a height of 9.1m, its velocity is (7.6i+6.1j)m/s. (i horizontal and j upward)
a) to what max. height does the ball rise?
b) what horizontal distance does the ball travel
(X-Xo=(Vo.COSangle)t right?)
c) what are the magitude and d) angle (below the horizontal) of the ball's velocity just before it hits the ground?


I did figure some parts out; since i is horiz. and j is upward the ball is still climbing. The magnitude of v is sq root of (7.6^2)+(6.1^2)=9.75 m/s and the angle at that moment is arctan(6.1/7.6) = 38.75degr.

I thought by calculating the horizontal range R I would get somewhere but I'm stuck again.
I think with the instantaneous velocity after 9.1m I must be able to calculate the t and the Vo and go from there but how?
Who can help me?

Answer 1

v = v0 + at. t it the time that the ball takes to get to v= 0 m/s
v0 = 6.1 m/s
a = -g = -9.8 m/s^2

So:

-9.8 t = -6.1

Here you calculate t

Then use the fact that the ball will be 2t in the air. Calculate then the x

You are right about the angle when the ball leaves the ground.

The formula that you put is right too.

To calculate the maximum you can use the formula

h = gt^2 / 2

Hope this helps. Contact me if you have more doubts

About the angle, consider the parabel that this ball draws.

Ana

Answer 2

Well, you know that acceleration is -g vertically (I'm going to guess you're ignoring air resistance, since I always do in my problems. If not, this is a little trickier). So using v squared = u squared + 2 * a * s, you can work out its initial velocity upwards -

v = 6.1
a = - g
s = 9.1

Use whatever value you've been taught for g (maybe 10, 9.8, or 9.81) Its initial horizontal velocity will just be 7.6. Go on from there - hope that helps!

Answer 3

You're lucky. They gave you the vector components ☺

So... V² = v0² - 2ax where v is final velocity, v0 is initial velocity, a is acceleration of gravity (9.8 m/s²), and x is distance traveled. Then (from the problem) 6.1² = v0² - 2*9.8*9.1
or v0 = √(6.1² + 2*9.8*9.1) = 14.682 m/s for the initial velocity. Since, at the top of it's path, the vertical component of the balls velocity is 0, we have
0 = 14.682² - 2*9.8*x or
x = 14.682²/(2*9.8) = 10.998 meters

How long does it take to fall 10.998 m?
s = at²/2 so t = √(2s/a) where s is distance.
t = √(2*10.998/9.8) = 1.498 s. It takes as long to get to the top as it takes to come down, so the total time in the air is 2*1.498 = 2.996 s. And, in that 2.996 s the horizontal distance travelled is 2.996*7.6 = 22.769 m.

The angle is arctan(j/i) = arctan(-14.682/7.6) = -62.63° and the magnitude is √(14.682² + 7.6²) = 16.532 m/s.


Doug

Answer 4

use the formula
v^2 -u^2= -2gh
0^2 - (6.1)^2 =-(2)(9.8)h
37.21=4.9 h
h=37.21/4.9=7.6 m..................i
Using h =ut + 1/2gt^2
t is the time when the ball will reach on the ground, h=0
0 = 6.1t + (0.5)(-9.8) t^2
0 = 6.1 - (4.9) t
t=6.1/4.9=1.224489796 seconds
Horizontal distance = 7.6x1.224489796=9.3meters..........ii

tan A =6.1/7.6=0.802631578
A= tan^-1(0.802631578)=43 degrees with horizontal............iii

Magnitude=sqrt{(6.1)^2 +(7.6)^2}=9.75m/s.......iv