Question

A boy throws a ball straight up into the air. It reaches the highest point of its flight after 4 seconds. How fast was the ball going when it left the boy’s hand? (a=10m/s2 )

Answer 1

40 m/s

Step-by-step explanation:

To solve this problem, we can use the following kinematic equation:

`\boxed{v = u + at}`,

where:

• v ⇒ final velocity = 0 m/s [as the ball stops moving when it reaches the highest point]

• u ⇒ initial velocity

• a ⇒ acceleration = -10m/s² [The value is negative as acceleration due to gravity is acting downwards but the ball is moving upwards]

• t ⇒ time taken = 4 s

Substituting the data into the equation above, we can calculate the initial velocity:

0 = u - 10 × 4

⇒ u = 40 m/s

Therefore, the ball was travelling at 40 m/s when it left the boy's hand.