Question

).A body is thrown with velocity of 40 m/s. It covers maximum horizontal distance of 60 m. Calculate maximum height reached.​

Answer 1

Explanation:

Correct option is C)

R.E.F image 

u=40m/s

θ=60m

g=9.81

Time of flight, T=g2usinθ

Maximum Height, H=2gu2sin2θ

Horizontal Range,  R=gu2sin2θ

Time taken to reach the 

max.height, t =2T

4=2×9.8(40)2(sin30∘)2=20.408m

=20.41m

R=9.8(40)2sin2×30∘=141.4m

t=2I=gusin3θ=2.041s

Ans =141.4m,20.41m,2.041s (C)