Question

A student throws a ball vertically up ward with a speed of 25m/s. The maximum height reached by the ball and its velocity respectively 5s after it was thrown is

Answer 1

Answer:

• h = 31.89 m
• velocity = -24 m/s

Explanation:

The maximum height reached by the ball can be calculated by ,

• h = v² - u² /2g

Here

• h is max. height
• v is final velocity (0 m/s)
• u is initial velocity (25 m/s)
• g is Acceleration due to gravity (9.8 m/s²)

Plugging the values,

• h = 0 - 25²/2*9.8
• h = 625/19.6
• h = 31.89 m

Hence, the maximum height reached by the ball is 31.89 m

The velocity can be calculated by,

• v = u + gt

where:

• v is final velocity (?)
• u is initial velocity ( 25 m/s)
• g is acceleration due to gravity (9.8 m/s²)
• t is time taken (5 seconds)

Plugging the required values,

• v = 25 - 9.8*5
• v = 25 - 49
• v = -24 m/s ( here negative sign indicates that the ball is moving in upward direction)