Question

A cannonball is fired from the ground at some angle and achieves a maximum height of 50.0 m and a horizontal range of 225 m. Find the following:
(a) Initial velocity.
(b) Angle of projection.
(c) Total time in the air.

Answer 1

Answer:

we know height H = u^2 sinα^2 / 2g

50 *2*10 = u^2 sinα^2

u^2 sinα^2 = 1000   .....1

also , Range R = u^2 sin2α  /g

u^2 sinα cosα = 225*10 = 2250  .....2

dividing 1 by 2 we have

tanα = 1000/2250 = .4 =

or α =  22°

from 1

u^2 = 1000/ sinα^2 = 1000 / 1000/.1369 =7304.6

u = 85.47

Time taken = 2u sinα /g = 2 (85.47) (.37) /10 = 6.324 s

Explanation: