A boy is standing on the top of a tower 100 m high. He throws a packet with a horizontal velocity of 10m * s ^ - 1 to his friend standing on the ground, 200 m away from the foot of the tower. Will the packet reach to his friend ? If not, how short will the packet full ?
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3
Answer:
From the figure, Tanθ=171/228
θ=Tan⁻¹[171/228]
The motion of projectile is from point A. Take reference axis at a.
θ=37° as u is below x axis.
u=15ft/s
g=32.2ft/s2
y=-171ft
From y= Tanθ-x²gsec²θ/2u²
-171=(tan37°)x- x²x32.2x(1.568)/2x225
-171=0.7536x- 0.1125x²
0.1125x²-0.7536x-171=0
on solving we can get x=35.78ft
Horizontal range covered by packet is 35.8ft .
So the packet will fall =228-35.8=192ft short of his friend.
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1
Answer:
35.8 feet is the answer
Explanation:
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