A target lies flat on the ground 7 m from the
side of a building that is 10 m tall, as shown
below.
The acceleration of gravity is 10 m/s
2
. Air
resistance is negligible.
A student rolls a 7 kg ball off the horizontal
roof of the building in the direction of the
target. The horizontal speed v with which the ball
must leave the roof if it is to strike the target
is most nearly what? (square roots)
To solve this problem, we can use the equations of motion in the vertical and horizontal directions.
First, let's consider the vertical motion of the ball. The time it takes for the ball to reach the ground can be found using the equation:
s = ut + (1/2)gt^2
where s is the vertical distance traveled (equal to the height of the building, 10 m), u is the initial vertical velocity (0 m/s since the ball is released from rest), g is the acceleration due to gravity (-10 m/s^2), and t is the time taken.
Plugging in the values, we get:
10 = 0 + (1/2)(-10)t^2
10 = -5t^2
t^2 = -10/5
t^2 = -2
t = √(-2) (Note: This is not a physically meaningful solution, as time cannot be negative)
Since we're interested in the horizontal speed with which the ball must leave the roof, we need to find the horizontal distance traveled by the ball in the same time t.
The horizontal distance traveled can be found using the equation:
s = ut
where s is the horizontal distance traveled, u is the initial horizontal velocity, and t is the time taken.
Since there is no horizontal acceleration (assuming no air resistance and neglecting the effect of gravity in the horizontal direction), the horizontal velocity remains constant throughout the motion.
Now, the horizontal distance traveled should be equal to the distance between the building and the target, which is 7 m. Therefore, we have:
7 = u√(-2)
To solve for u, we need to get rid of the square root. Squaring both sides of the equation, we get:
49 = u^2(-2)
u^2 = -49/-2
u^2 = 24.5
u = √24.5
So, the horizontal speed with which the ball must leave the roof to strike the target is most nearly √24.5 (square root of 24.5), or approximately 4.95 m/s.