Question

An object is projected at an angle of 600 with the horizontals. If the horizontal range of

the object is 2.5 km, make calculation for velocity of projection and maximum height

attain by the object.​

Answer 1

Answer:

Let the velocity of projectile be u.

Angle of projection with the horizontal θ=60

o

So, initial velocity in horizontal direction u

x

=ucosθ=u×cos60

o

=0.5u

Horizontal distance covered in 5 seconds is 90 m

Thus 0.5u=

5

90

⟹ u=36 m/s

Now the body is projected vertically upwards with a speed u=36 m/s

Maximum height reached H=

2g

u

2

=

2×10

36×36

=64.8 m

Explanation:

hope its helpful