Question

If a ball is thrown up at an initial speed of 40. m/s, how many seconds does it take to reach the top of its path?

Answer 1

Time, t = 4.08 secs

Step-by-step explanation:

Given the following data;

Initial velocity, U = 40m/s

To find the time, we would use the first equation of motion;

`V = U + at`

Where;

• V is the final velocity.

• U is the initial velocity.

• a is the acceleration.

• t is the time measured in seconds.

Making time, t the subject of formula, we have;

`t = (V - U)/(a)`

We know that acceleration due to gravity, g is 9.8m/s².

a = g = - 9.8m/s² because the ball is thrown in the opposite direction.

Also, the final velocity is equal to zero (0) because the ball reached its maximum height.

Substituting into the equation, we have;

`t = (0 - 40)/(-9.8)`

`t = (-40)/(-9.8)`

Time, t = 4.08 secs

Therefore, it will take the ball 4.08 seconds to reach the top.