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?

h = 0.5g*t^2 = 100-25

4.9*t^2 = 75
t^2 = 15.31
t = 3.91 s. = Fall time.

Dx = Xo*t = 20 m.
Xo*3.91 = 20
Xo = 5.11 m/s. = Hor. velocity.

5.0s

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?

To find the horizontal velocity at which the ball should be imparted, we can use projectile motion equations.

The first step is to determine the time it takes for the ball to travel from the first tower to the second tower. We can use the formula for the time of flight in vertical motion:

t = √(2h/g)

Where:
t = time of flight
h = vertical distance (in this case, the height of the first tower - the height of the second tower)
g = acceleration due to gravity (approximately 9.8 m/s^2)

Substituting the values, we get:

t = √(2(100 - 25)/9.8)
= √(150/9.8)
≈ √15.31
≈ 3.91 seconds (rounded to two decimal places)

Now that we know the time of flight, we can determine the horizontal velocity. The horizontal velocity remains constant throughout the motion as there is no horizontal acceleration. The formula for horizontal velocity is:

v = d / t

Where:
v = horizontal velocity
d = horizontal distance (in this case, the distance between the two towers)
t = time of flight

Substituting the values, we get:

v = 20 / 3.91
≈ 5.11 m/s (rounded to two decimal places)

Therefore, the ball should be imparted with a horizontal velocity of approximately 5.11 m/s in order to land on the rooftop of the second building.