Question

A body is thrown up with a velocity of 78.4 m per second.How high will it rise and how much time it will take to return to its point of projection

Answer 1

The maximum height reached by the body is 313.6 m

The time to return to its point of projection is 8 s.

Step-by-step explanation:

Given;

initial velocity of the body, u = 78.4 m/s

at maximum height (h) the final velocity of the body (v) = 0

The following equation is applied to determine the maximum height reached by the body;

v² = u² - 2gh

0 = u² - 2gh

2gh = u²

h = u²/2g

h = (78.4²) / (2 x 9.8)

h = 313.6 m

The time to return to its point of projection is calculated as follows;

at maximum height, the final velocity becomes the initial velocity = 0

h = v + ¹/₂gt²

h = 0 + ¹/₂gt²

h = ¹/₂gt²

2h = gt²

t² = 2h/g

`t = \sqrt{(2h)/(g) } \\\\t = \sqrt{(2* 313.6)/(9.8) }\\\\t = 8 \ s`