Question

A ball is thrown upward at time t=0 from the ground with an initial velocity of 8 m/s (~ 18 mph). What is the total time it is in the air? Assume that g = 10 m/s2.

Answer 1

Answer:

The total time it is in the air for the ball is 1.6326 s

Given:

Initial velocity = 8 $\frac{m}{s}$

To find:

the total time it is in the air = ?

Formula used:

t = $\frac{v-u}{a}$

Where t = time to reach maximum height

v = final velocity of the ball = 0 m/s

u = initial velocity of ball = 8 m/s

a = acceleration due to gravity = -9.8

Acceleration of gravity is taken as negative because ball is moving in opposite direction.

Solution:

A ball is thrown upward at time t=0 from the ground with an initial velocity of 8 m/s.

The time taken by the ball to reach the maximum height is given by,

t = $\frac{v-u}{a}$

Where t = time to reach maximum height

v = final velocity of the ball = 0 m/s

u = initial velocity of ball = 8 m/s

a = acceleration due to gravity = -9.8

Acceleration of gravity is taken as negative because ball is moving in opposite direction.

t = $\frac{0-8}{-9.8}$

t = 0.8163 s

Thus, time taken by the ball to reach the ground again = time taken to reach maximum height

So, Total time required for ball to reach ground = 2t = 2 × 0.8163

Total time required for ball to reach ground = 1.6326 s

The total time it is in the air for the ball is 1.6326 s