Question

Rodger decides to pass the time by throwing a baseball straight up in the air. He measures that it takes the ball 1.05 seconds to return from its maximum height to its starting height. How high above the ground was the baseball when it reached its maximum height?

Answer 1

5.40 m to 3 sig figs

Explanation:

We'll assume metric units, since none are specified. The equation for predicting the distance and time of a free-falling object on Earth is given by:

y = y₀ + v₀t - (1/2)gt^2 where

y = distance (m)

y₀ is the starting height (m)

v₀ is the initial velocity

t is time in seconds, and

g is the acceleration due to gravity (-9.8 m/s^2)

In our case, y is distance above Earth, which will be 0 meters (it returns to Earth).

y₀ is the starting height (m) which is the unknown in this case.

v₀ is the initial velocity, which is zero since it reached its maximum height and has slowed to zero m/s.

t is time in seconds, 1.05 seconds here.

Plug in the values:

y = y₀ + v₀t - (1/2)gt^2

0 = y₀ + (0)*t - (1/2)(9.8 m/s^2)*(1.05 s)^2

-y₀ = - (1/2)(9.8 m/s^2)*(1.1025 s^2)

-y₀ = - (1/2)(9.8 m/s^2)*(1.1025 s^2)

-y₀ = - (-5.40225 m)

y₀ = 5.40225 m

y₀ = 5.40 m to 3 sig figs