Question

A particle is projected with velocity u' at an angle of elevation 55 with horizontal. It reaches to the height of 4.8 m in 3 sec. Determine velocity u' and the range.

Answer 1

Answer:

We have been given the max.height reached by the body,

H = 4.8 m

Time taken to cover H = t = 3 s

angle of elevation = theta = 57°

therefore, time taken to cover entire journey = 2 × t = 2 × 3 = 6s

Using the formula for time of flight of a projectile -:

T = 2u sin theta/g

6 = (2u sin 57)/10

(taking g = 10m/s^2)

6/2 = u × (4/5)/10

3 = 4u/5×10

3 × 50/4 = u

u = 150/4 = 37.5 m/s

Now, range = u^2 sin 2 × theta/g

R = (37.5)^2 × sin (2 × 57)° / 10

R = 1406.25 × 2 × sin theta × cos theta / 10

(sin 2 theta = 2 sin theta cos theta)

R = [1406.25 × 2 × (4/5) × (3/5)]/ 10

R = (1406.25 × 24/25)/10

R = 1350/10 = 135 m

Hence the projectile covers R = 135 m and its initial velocity, u = 37.5 m/s