An astronaut tosses a ball upwards on an unknown planet with no atmosphere. The astronaut gives the ball a speed of 5.0 m/s. It takes 5.0 seconds to return to the height it was tossed from.

Question

1. What is the acceleration of gravity on this unknown planet?
2. How high did the ball rise above the astronaut’s hand?

Answer 1

Given

Point A

Vo = 5 m/s

t = 5 seconds (to return to same height, let this be point B)

Find

1) a of gravity ?

2) hmax = ?

Solution

1)

Motion from A to B

Let the origin be the point where the ball is tossed upward

Y = -(1/2) * g * t^2 +Voyt + Yo

Yo = 0

Y = 0

A and B are at the same height level, given on problem statement

0 = -(1/2) * g * t^2 + Voyt

g = 2 * Voy / t = 2 * 5 /5 = 2 m/s^2

g = 2 m/s^2 (downward)

2) Let C be the point of maximum height

Vyc ^2 = 2a(Yc-Ya)+Vya^2

Vya = Vo = 5 m/s

Vyc = 0 (At maximum height velocity is zero)

0 = -2*g*Yc + Vya^2

2gYc = Vya^2

hmax = Yc = Vya^2/(2*g) = 5^2/(2*2) = 6.25 m

Answer 2

v_f = v_i + at

If the ball goes up to the halfway point (top of the flight), then at that point its velocity will be zero (occurring at t = 2.5 s).

So 0 = 5 + a(2.5)

-5 = 2.5a

a = -5/2.5 = -2

the acceleration due to gravity = -2 m/s²

The height of the ball is calculated from x = v_it + 1/2at²

x = (5)(2.5) + 1/2(-2)(2.5)²

x = 6.25 meters