Question

A ball is at 35 m horizontal distance from the foot of 10m high building as shown in figure. The ball is
projected with a velocity of 25 m/s at an angle of 37° with horizontal. At how much distance (in meters)
from the corner point A, will the ball hit the roof ? (Assume that roof is sufficiently large so that ball lands
on it. Neglect air resistance, take g = 10 m/s)
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Answer 1

Answer:

Range =u^2sin2theta/g....formula

R=625*2*3/5*4/5/10=60meter.

H=625*9/20*25=45meter.

Therefore the ball will not reaxhes at the top of the building.

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Answer 2

Answer:

The ball hit roof at 58 m from point A.

Explanation:

Given that,

Distance = 35 m

Height = 10 m

Velocity = 25 m/s

Angle = 37°

We need to calculate the distance from the corner point A

Using equation of trajectory

$y = x\tan\theta-\dfrac{gx^2}{2v^2\cos^2\theta}$

Put the value into the formula

$10=x\tan37-\dfrac{10\times x^2}{2\times(25)^2\cos37}$

$x = 17,58$

The roof is starting 35 m from point A.

So the value of x less than 35 m will be ignore.

Hence, The ball hit roof at 58 m from point A.