Question

A body is projected with velocity of 40m/s after 2s it crosses a vertical pole of height 20.4m calculate the angle of projection and horizontal range.

Answer 1

u = 40 m/s     g = 9.8 m/s

At   t1 = 2 sec    y = 20.4 m
y = u sinФ  t  - g t² /2 
20.4 = 40 sinФ * 2 - g 2² /2
=>  sinФ = 1/2
=>  Ф = 30°

Horizontal range = R = u² Sin2Ф / g = 40² sin60° / 9.8  = 141.39 m

Answer 2

Answer:

30 DEGREE , 141.4 M

Explanation : u = 40 m/s     g = 9.8 m/s

At   t1 = 2 sec    y = 20.4 m

y = u sinФ  t  - g t² /2 

20.4 = 40 sinФ * 2 - g 2² /2

=>  sinФ = 1/2

=>  Ф = 30°

Horizontal range = R = u² Sin2Ф / g = 40² sin60° / 9.8  = 141.39 m

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