A ball is shot from the ground straight up into the air with initial velocity of 46ft/sec. Assuming that the air resistance can be ignored, how high does it go?

Question

Hint: The acceleration due to gravity is 32 ft per second squared.

Answer 1

From Physics, we will use the kinematics equation: v² = v_o² + 2a(x - x_o) where v is the final velocity, v_o is the initial velocity, a is the acceleration, x is the final height, and x_o is the initial height.

We are looking the final height, so we will solve for x in the above kinematics equation:

v² = v_o² + 2a(x - x_o)

v² - v_o² = 2a(x - x_o)

(x - x_o) = v² - v_o² / 2a

x = v² - v_o² / 2a + x_o

At the top of the ball's height, it's final velocity will be 0. Also, we will set the initial height at 0 ft. In this case, the initial velocity is 46 ft/sec, and the acceleration is -32 ft/sec/sec (negative because gravity working against the ball's velocity).

Plugging in, we get:

x = ((0)² - (46)²) / (2(-32)) + 0 = 62.125 ~= 62.2 ft.