Question

A ball is Thrown Upward with An initial velocity of 30 m/s , How high does it rise from the ground when the ball reaches the highest Point ? And how long does it take to fall down to half of its height

Answer 1

Answer:

• The maximum height reached by the ball is 45.92 m

• Time taken to fall down to half of its height is 2.2 s

Explanation:

Given;

initial velocity of the ball, u = 30 m/s

final velocity of the ball at the highest point, v = 0

The maximum height reached by the ball is calculated as;

v² = u² - 2gh

where;

h is the maximum height reached by the ball

0 = 30² - (2 x 9.8)h

19.6h = 900

h = 900 / 19.6

h = 45.92 m

Time taken to fall to half of its height is calculated as;

when falling down, the final velocity v becomes the initial velocity = 0.

Apply the following kinematic equation;

h = ut + ¹/₂gt²

h = 0 + ¹/₂gt²

h = ¹/₂gt²

where;

h = 45.92 m is the maximum height reached

half of h = 45.92 / 2 = 22.96 m

22.96 = ¹/₂gt²

$t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 22.96}{9.8} }\\\\t = 2.2 \ s$