Question

body is projected with a velocity of 20m/s at an angle of 60degree with the vertical. Find the maximum height, time of flight and the horizontal range.​

Answer 1

Answer:

Given, u=20ms−1,θ=60∘,t=0.5s

(i) Since, horizontal distance,

x=(u cosθ)t=(20cos60)×0.5=5m

Similarly, vertical distance,

y=(u sinθ)t−12gt2

=(20sin60∘)×0.5−12×9.8×(0.5)2⇒y=7.43m

(ii) Velocity along horizontal direction,

vx=ucosθ=20cos60∘=10ms−1

Velocity along vertical direction,

vy=usinθ−gt=20sin60∘−9.8×0.5=12.42ms−1