A boy stands on the roof of a 50m tall building and thrown a ball horizontslly at 20m/s away from the building.
How far away from the foot of the building will the ball land?
the free fall equation is
... h = -4.9 t^2 + 50
set h = 0 and solve for the flight time
multiply the flight time by 20 m/s to find the impact spot
I got an answer of 63.8m is it correct
to three figures ... 63.9
do you use significant figure rules?
To determine how far away from the foot of the building the ball will land, we can use the projectile motion equations. Since the initial velocity of the ball is only in the horizontal direction, there is no vertical component initially.
First, let's find the time it takes for the ball to reach the ground. We can use the equation:
h = (1/2) * g * t^2
where h is the initial height (50m), g is the acceleration due to gravity (9.8 m/s^2), and t is the time.
Rearranging the equation, we have:
t = sqrt((2 * h) / g)
t = sqrt((2 * 50) / 9.8)
t ≈ 3.19s
Now, we can find the horizontal distance covered by the ball using the equation:
d = v * t
where d is the distance, v is the initial horizontal velocity (20 m/s), and t is the time.
d = 20 * 3.19
d ≈ 63.8m
Therefore, the ball will land approximately 63.8 meters away from the foot of the building.