Question

the ceiling of a long hall is 25m high. what is the maximum horizontal distance that a ball thrown with a speed of 40 m/s can go with other hitting the ceiling fan


ans = 150.5
I need steps ​

Answer 1

Answer:

Explanation:

Let us consider that the ball is thrown at an angle θ

with horizontal.

We know the formula for calculating the maximum height the ball has attained-

H=u2sin2θ2g

Where , u

is the speed of ball, θ

is the angle of projection with horizontal and g

is the acceleration due to gravity having value 9.8m/s2

.

Therefore 25=(40)2sin2θ2×9.8

Rearranging the terms sin2θ=0.30625sinθ=0.5534

The value of θ

is found to be θ=sin−1(0.5534)θ=33.60∘

So with angle θ=33.60∘

, the ball will hit the maximum height of the hall.

Now applying formula for horizontal range,

R=u2sin2θg=(40)2sin(2×33.60∘)9.8=1600sin(67.2)9.8=150.53m

Therefore the range is found to be 150.53m.

Thus the ball can be thrown at a distance of 150.53m

Additional information: The range of an object is the maximum horizontal distance the object can cover when projected at any angle. Similarly, maximum height is referred to as the greatest height the object projected can attain.

Note: The unit of the parameters used should always be kept in mind while using formulas. Here the velocity is in m/s

and acceleration due to gravity in m/s2

. The formula for height and range looks alike as there is a minute difference therefore it should be used carefully.