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Grade: 11
        A ball is projected from a point on the horizontal ground at horizontal distance of 39.2m from the foot of a building. When it is at the highest point of its trajectory , it passes through a window of the building,which is at a vertical height of 19.6m above ground. Calculate the speed of projection.g=9.8m/s
2 years ago

Answers : (1)

Agrata
13 Points
							
Let The velocity of projection be u and angle be ∂
Max height attained is 19.6m
At highest point vy=0 thus from third equation of motion
Vy2=uy2-2gh
0=(uy)2-2*9.8*19.6
Uy=(2*9.8*19.6)1/2=19.6m/s
Fron second equation of motion
H=uyt-1/2gt2
19.6=19.6t-4.9t2
t=2sec
Max horizontal distance covered is 39.2m
39.2=ux*2
ux=39.2/2=19.6m/s
u=(ux2+uy2)1/2=19.6(2)1/2
Since ux=uy angle of projection is 45deg
2 years ago
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