Question

An object is projected from a height of 100m above the ground at an angle of 300to the horizontal with a velocity of 100m/s.
Calculate
(i) The maximum height reached above the ground (4)
(ii) Time of flight (4)
(iii) The velocity and the direction of the object 1 sec before it hit the ground (4)

Answer 1

Explanation:

Given,

Initial velocity =100m/s

Angle =30

0

So,

The initial velocity makes an angle 30° with horizontal. So,its vertical component will be 100sin30

0

=50m/s.

The time taken by the ball to acheive max height will be at a point when its vertical velocity will be zero. So Vsinx=at

T=

a

vsinx

=5seconds.

So after 5 seconds it will reach its max height.

Following the symmetry,it will land after 10seconds.

So the total horizontal distance covered will be vcosx×10≈866m

.I hope that is helpful for u.

thanks