A ball is released from a tower at a height of 100 meters toward the roof of another tower that is 25 meters high. The horizontal distance between the two towers is 20 meters. With what horizontal velocity should the ball be imparted so that it lands on the rooftop of the second building?

Hint:

Calculate the time it takes to reach the lower roof, say t (in seconds), by solving for t in:
(100-25)=(1/2)gt²

The minimum horizontal velocity will be 20/t m/s

To find the horizontal velocity with which the ball should be imparted, we can use the principle of projectile motion. The ball is released from a height of 100 meters and needs to land on the rooftop of the second building that is 25 meters high. The horizontal distance between the two towers is 20 meters.

Projectile motion can be broken down into horizontal and vertical components. In this case, we need to find the horizontal velocity, so we'll focus on the horizontal component.

First, let's find the time it takes for the ball to fall from a height of 100 meters to a height of 25 meters. We can use the equation for vertical displacement in free fall:

y = ut + (1/2)gt^2

Where:
y = vertical displacement (difference in height between the two towers)
u = initial vertical velocity (0 m/s as the ball is released from rest)
g = acceleration due to gravity (approx. 9.8 m/s^2)
t = time

Rearranging the equation, we have:

y = (1/2)gt^2

Plugging in the values:
75 = (1/2)(9.8)t^2

Simplifying further:
75 = 4.9t^2

Dividing both sides of the equation by 4.9:
t^2 = 15.306122

Taking the square root of both sides:
t ≈ 3.91 seconds (rounded to two decimal places)

Now that we have the time it takes for the ball to fall, we can calculate the horizontal velocity using the equation for horizontal displacement:

x = ut

Where:
x = horizontal displacement (20 meters, distance between the two towers)
u = horizontal velocity (what we need to find)
t = time (3.91 seconds)

Rearranging the equation and substituting the known values, we have:

20 = u × 3.91

Solving for u:
u = 20 / 3.91

Calculating this:
u ≈ 5.12 m/s (rounded to two decimal places)

Therefore, the ball should be imparted with a horizontal velocity of approximately 5.12 m/s to land on the rooftop of the second building.