Question

1. Particle is thrown with speed of 20 m/s at angle of
projection 53° with the horizontal find a) time of ascend
b) time of flight c) maximum height d) horizontal range.​

Answer 1

Answer:

Solution



Verified by Toppr

Correct option is

D

5m

Given :  u=20 m/s     θ=30o      g=10 m/s2

We get  sinθ=sin30o=21

Maximum height reached   H=2gu2sin2θ=2020×20×(21)2=5m

HOPE IT WORKS