Question

e.5 Maximum height of a projectile is 30% less than its
horizontal range. Calculate its angle of projection.
solution plz.​

Answer 1

Answer:

513ms−1

At the top point, horizontal displacement is equal to half the range, vertical displacement is equal to the maximum height of the projectile and time taken is half the time of flight.

Horizontal displacement is:

x=2R=2u2sin(2θ)/g 

x=2×10202sin(60o)=103 m

Vertical displacement is:

y=H=2gu2sin2θ

y=2×10202sin2(300)=5 m

Magnitude of total displacement is:

s=x2+y2

s=(103)2+52

s=513 m

Time taken is:

t=2T

t=22usinθ/g

t=