Can anyone tell me what equation to use to solve this problem? I am having problem solving it. Thanks?

Question

A ballplayer catches a ball 3.0s after throwing it vertically upward. With what speed did he throw it and what height did it reach?

Answer 1

this is a motion problem and there are four possible equations you need to know for these types of problems

(1) Vf = Vi + at
(2) d = 1/2 (Vf + Vi) x t
(3) d = Vit + 1/2 a t^2
(4) Vf^2 = Vi^2 + 2ad

where Vf = velocity final, Vi= velocity initial, a = acceleration, t = time, d = distance

now to solve this problem think of this. you throw the ball up. it slows down until it reaches it's maximum height. then stops for an instant then falls again. in this motion, if it starts with a velocity "v" upward, when it returns to the same position as it started, it's velocity will equal "v" except in the opposite direction. Also, the time up = time down...

so what you have is a ball going up, stopping, then coming down.

total time = time up + time down = 3.0 sec
time up = 3.0 sec/2 = 1.5 sec

so you know "t", "Vf" = 0 at the top, and you need to know "d" and "Vi"

from equation (1) above

Vf = Vi + at
Vi = Vf-at = 0 - (-9.8 m/s^2) x (1.5 s) = 14.7 m/s

fyi, a = 9.8 m/s^2 downward = -9.8 m/s^2 up (acceleration due to gravity). in ft/s^2, it's 32 ft/s^2...

from equation (2) above...

d = 1/2 (Vf + Vi) x t = 1/2 (0m/s+14.7m/s) x (1.5s) = 11 m

or if you prefer, from equation (3) above

d = Vit + 1/2 a t^2
d = (14.7 m/s) x (1.5s) + 1/2 x (-9.8 m/s^2) x (1.5s)^2
d = 11 m

Answer 2

First consider the displacement. Since the ball reaches the point from where it starts, displacement, s = 0

We know that s = ut + 1/2at^2

Therefore, 0 = -u * 3 + 0.5 * 9.8 * 3 * 3 (-u because its going upwards first)

Solving the equation, we get

u = 14.7 m/s

So the speed at which he throws the ball upwards is 14.7 m/s
*****

Now to calculate the height:

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

v = 0, because the ball comes to a stop at the instant it reaches the highest point (at t=1.5)

so,

- u*u = 2 * a * s

- 14.7*14.7 = 2 * 9.8 * s

Solving for s,
we get

s = 10.8 m
*****

Hope this helps!

Therefore, the highest point the ball reaches is 10.8 m.

Answer 3

The first thing to solving this problem is realizing the ball goes up and then falls back down, so that means it travels UP 1/2 the time (of 3.0s) and DOWN 1/2 the total time. This should make it easy to use your formula for acceleration of falling objects due to gravity. Does 32 feet per second per second sound familiar? I thought so. Now solve your problem. You can do it!

Answer 4

s=V0*t - (1/2)*9.8*t²

HINT: When the ball gets back to the ground, s = 0 and the value of t that is half that is the t value at it's maximum heigth.

HTH

Doug