Question

1. Hoai Nguyen, a Physics 2A student, drop a soccer ball from the roof of the new science building. The ball strikes the ground in 3.30 s later. You may ignore air resistance, so the ball is in free fall. How tall, in meters, is the building?
2. How fast was the ball moving right before hitting the floor?

3. Thu Tran, another Physics 2A student, grabs the ball and kicks it straight up to Hoai Nguyen, who is still up on the building rooftop. Assuming that the ball is kicked at 0.50 m above the ground and it goes on a vertical path, what is the minimum velocity required for the ball to make it to the building rooftop? Ignore air resistance. (Hint: the ball will pass the rooftop level with a higher speed...)

Answer 1

Answer:

1. 53.415 m

2. 32.373 m/s

3. 30.82 m/s

Explanation:

Let g = 9.81 m/s2. We can use the following equation of motion to calculate the distance traveled by the ball in 3.3s, and the velocity it achieved

1.$s = gt^2/2 = 9.81*3.3^2/2 = 53.415 m$

2.$v = gt = 9.81*3.3 = 32.373 m/s$

3. If the ball is kicked at 0.5 m above the ground then the net distance between the ball and the roof top is

53.415 - 0.5 = 48.415 m

For the ball to at least make it to the roof top at speed v = 0 m/s. We can use the following equation of motion to calculate the minimum initial speed

$v^2 - v_0^2 = 2g\Delta s$

where v = 0 m/s is the final velocity of the ball when it reaches the rooftop, $v_0$ is the initial velocity, $\Delta s = 48.415$ is the distance traveled, g = -9.81 is the gravitational acceleration with direction opposite with velocity

$0 - v_0^2 = 2*(-9.81)*48.415$

$v_0^2 = 950$

$v_0 = \sqrt{950} = 30.82 m/s$