# The time of flight of a projectile is 10 sec.its horizontal range is 100m.Calculate angle and velocity of projection

Chetan Mandayam Nayakar
312 Points
13 years ago

v2sin(2theta)=100g, vsin(theta)=5g,thus v2=25g2/sin2(theta), thus 50g2cot(theta)=100g, theta=arccot(1/5), v=10√26 m/sec

509 Points
13 years ago

time of flight=2usin@/g=10........1

horizontal range=u^2sin2@ /g=100........2

dividing 1by2

usin2@/2sin@=10 or

ucos@=10....3

divide 1 by 3

we get tan@=5 ,cos@=1/sqrt26

on putting value of @ in 3 we get

u=10sqrt26

Ashwin Sheoran
44 Points
5 years ago

Let θ be the angle of projection (with the horizontal) and ‘u’ be the velocity of projection. Given time of flight is, t = 10 s.

Horizontal range, x = 100 m

=> u cosθ × 10 = 100

=> u cosθ = 10 …………………..(1)

=> u2 cos2θ = 100 ………………..(2)

Now, considering the vertical motion of the projectile,

v = u + at

=> -u sinθ = u sinθ – gt (since, initial vertical velocity = u sinθ, final vertical velocity after time t is –u sinθ)

=> u sinθ = gt/2 = (10 × 10)/2 = 50 ………………..(3)

=> u2 sin2θ = 2500 ………………………………….(4)

(2) + (4) => u2 = 2600

=> u = 50.99 m/s

(3)/(1) => tanθ = 5

=> θ = 78.690