Question

A ball is thrown up into the air with the initial velocity of 18 m/s. A) How high does the ball go? B) Calculate the time needed for the ball to reach its max height.

Answer 1

Answer:

(a) h = 16.53 m (b) t = 1.83 s

Explanation:

Given that,

The initial velocity of a ball, u = 18 m/s

When it reaches to the maximum height, its final velocity v will be 0. Let it goes to a maximum height of h meters.

Finding t using first equation of motion as follows :

v = u +at

Here, a = -g and v = 0

$t=\dfrac{u}{g}\\\\t=\dfrac{18}{9.8}\\\\t=1.83\ s$

The time needed for the ball to reach its max height is 1.83 s.

Let h is the maximum height. Using second equation of motion to find it :

$h=ut-\dfrac{1}{2}gt^2\\\\h=18(1.83)-\dfrac{1}{2}\times 9.8\times (1.83)^2\\\\h=16.53\ m$

So, it will go to a maximum height of 16.53 m.