physics question!3?

Question

1.) You stand at the top of a cliff while your friend stands on the ground below you. You drop a ball from rest and see it takes 1.2 secs. for the ball to hit the ground below. Your friend then picks up the ball and throws it up to you, such that it just comes to rest in your hand. What is the speed with which your friend threw the ball.

Answer 1

a=vf-vi/t > vf-vi=a*t

Vf=0
0-Vi= -9.98*1.2=11.976m/s

Answer 2

For this problem, three equations are needed. First, average velocity (v.avg) = initial velocity (v.i) + final velocity (v.f) / 2. Second, v.avg = change in position (that is, final position - inital position, or p.f - p.i) / change in time (t.f - t.i). Third, acceleration (a) = change in velocity (v.f - v.i) / change in time.

The problem must be solved in two parts. We know that final position is 0 m (the ground), initial velocity is 0 m/s (the ball is dropped from rest), change in time = 1.2 s, and acceleration = -9.80 m/s^2 (because the ball's motion is not in a vacuum, acceleration is due to gravity; it is negative because gravity attracts objects on Earth downward to the core...down is the "negative" direction, while up is "positive."). However, we do not know initial position or final velocity.

Because a = (v.f - v.i) / t and v.i = 0, a = v.f / t. Rearrange the equation to solve for v.f (the speed at which the ball reaches your friend and at which your friend throws it up to you): v.f = a * t = (-9.80 m/s^2) * (1.2 s) = -11.76 m/s (if you like, use significant digits, or -12 m/s; for the sake of precision, I'll use it without). Because we now know v.f, we can solve for the change in position: (v.i + v.f) / 2 = (change in position) / t; v.i is 0 m/s, so it can be eliminated from the equation. Rearrange: change in position = (v.f * t) / 2 = (-11.76 m/s * 1.2 s) / 2 = -7.056 m, and p.i = 7.2 m (because change in position = p.f - p.i and p.f = 0 m).

All right. Whew. Second part now; although the speed should be the same as v.f in part one, let's check. We know that change in position = 7.2 m (positive now because the ball is moving up), a = -9.80 m/s^2, t.i = 0 s and v.f = 0 m/s. However, v.i (the speed at which your friend throws the ball) and t.f are unknown. Thus we have two equations: (v.i + v.f) / 2 = (change in position) / (t.f - t.i) and a = (v.f - v.i) / (t.f - t.i). Since v.f is 0 m/s and t.i = 0s, eliminate them from the equation. Solve for t.f in both equations: first, t.f = 2(change in position) / v.i and second, t.f = -v.i / a. Now substitute so that 2 change in position) / v.i = -v.i / a and solve for v.i. We have -(v.i)^2 = 2 * change in position * a. Take the square root of both sides; it must be positive because the ball is being thrown up, so v.i = (the square root of) 2 * change in position * a = (the square root of ) 2 * 7.056 m * -9.80 m/s^2 (ignore the negative sign here, take the square root without it). So v.i = 11.76 m/s = the same speed (except in a positive direction) at which you threw the ball down.

There. I hope it helped. :)

Answer 3

using v = ut + 1/2at^2
rearrange to find u
u = (v/t) - 1/2at
= (0/1.2) - 1/2 x -9.8 x 1.2
= 5.88m/s

Answer 4

Using the formula when ball is thrown down with initial velocity u=0, g=9.8 t=1.2 seconds and h=height
h=ut+1/2 gt^2
h=0(t) +0.5 (9.8)(1.2)^2
h=7.056 m...........................i
When ball is thrown upwards us the formula v^2 -u^2 = -2gh
here v=0 and u to find h=7.056 meter
0^2 -u^2 = -2(9.8)(7.056)
u^2 = 34.5744
Taking sq rt
u=5.88 m/sec.......................ii

Answer 5

(1.2)(9.8)=11.76 m/s