A ball thrown vertically upward is caught by the thrower after 1.63 s. Find the maximum height the ball reaches. he acceleration of gravity is 9.8 m/s

Answer 1

Ball takes the same amount of time to go up as it does to come down, meaning that it takes 1.63/2 = 0.815 seconds to reach the top. Additionally, we know that the velocity of the ball at the top is 0, and that it experiences a constant acceleration of 9.8 m/s² in the downward direction.

Now, we use the equation v_f = v₀ + at as well as the equation v_f² = v₀² + 2ah.

We take v_f to be the velocity at the top of the trajectory (0 m/s), v₀ as the velocity of the ball when thrown (unknown), t as the time to reach the top of the trajectory (0.815 s), and a as the acceleration due to gravity (-9.8 m/s²; negative because it acts in the direction opposite to the velocity).

Plugging into the first equation of v_f = v₀ + at, we get:

0 = v₀ - 9.8*0.815

v₀ = 7.987

Plugging into the second equation of v_f² = v₀² + 2ah, we get:

0² = 7.987² - 2*9.8*h

h = 3.2547025 m

This tells us that the max height reached by the ball is about 3.25 meters.

Hope this helps!