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)
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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
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