A target lies flat on the ground 4 m from the side of a building that is 10 m tall, as shown below.
The acceleration of gravity is 10 m/s2 . Air resistance is negligible.
A student rolls a 4 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
To find the horizontal speed (v) with which the ball must leave the roof in order to strike the target, we can use the equations of projectile motion.
First, let's consider the motion in the vertical direction. We know that the height of the building is 10 m, so the time taken for the ball to fall from the roof to the ground can be found using the formula:
h = (1/2)gt^2
where h is the height (10 m), g is the acceleration due to gravity (10 m/s^2), and t is the time.
Rearranging the equation, we have:
t^2 = (2h)/g
t^2 = (2*10)/10
t^2 = 2
t = √2
Now, let's consider the motion in the horizontal direction. The horizontal distance that the ball must cover is 4 m. The time taken for the ball to reach this distance can be found using the formula:
d = vt
where d is the distance (4 m), v is the horizontal velocity, and t is the time.
Rearranging the equation, we have:
v = d/t
v = 4/√2
v = 2.83 m/s
Therefore, the horizontal speed (v) with which the ball must leave the roof in order to strike the target is most nearly 2.83 m/s.
To solve this problem, we can use the equations of motion to find the required horizontal speed of the ball.
First, let's identify the information given in the problem:
- The height of the building (h) = 10 m
- The distance between the target and the building (d) = 4 m
- Acceleration due to gravity (g) = 10 m/s^2
We need to find the horizontal speed (v) with which the ball must leave the roof to hit the target.
To start, we can use the equation of motion for vertical displacement:
h = (1/2)gt^2
Since the ball is dropped from rest on the roof, the vertical displacement is equal to the height of the building. So we can rewrite the equation as:
10 = (1/2)(10)t^2
Simplifying, we have:
t^2 = 2
t ≈ √2
Now, we can find the time it takes for the ball to reach the target horizontally. Since there is no horizontal acceleration, the time taken for the horizontal distance is the same as the time taken for the vertical drop.
t ≈ √2
Finally, we can find the horizontal speed needed to travel the 4 m distance:
v = d / t
v = 4 / (√2)
v ≈ 2.83 m/s
Therefore, the horizontal speed (v) with which the ball must leave the roof to hit the target is most nearly 2.83 m/s.
How long to fall 10 meters?
10 = (1/2)(10) t^2
solve for t, time in air
then
4 meters = u t
solve foe u