A playground is on the flat roof of a city school, 5.2 m above the street below (see figure). The vertical wall of the building is h = 6.50 m high, forming a 1.3-m-high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of è = 53.0°; above the horizontal at a point d = 24.0 m from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall.
a) Find the speed at which the ball was launched.m/s
b) Find the vertical distance by which the ball clears the wall. m
c Find the horizontal distance from the wall to the point on the roof where the ball lands. m
I forgot to add the questions to my question earlier. Here they are now. Thanks for your help.
a) The horizontal component of the launch speed (Vo) is Vo cos = 0.6018 V. It travels 24 m horizontally in 2.20 s. Therefore
0.618 Vo * 2.2 = 24.0. Solve for Vo
Vo = 17.65 m/s
The vertical lauch speed component is Vo sin 53 = 14.10 m/s. Use the equation of vertical motion,
y = (Vo sin 53)* t - (1/2) g t^2
to solve for the height after t = 2.2 s
c) Use the same equation to solve for t when y = 5.2 m
To solve this problem, we will need to use the equations of motion for projectile motion. Here's how you can find the solutions for each question:
a) To find the speed at which the ball was launched, we need to find the initial velocity of the ball. The given information tells us that the ball takes 2.20 s to reach a point vertically above the wall (this is the time it takes for the vertical motion). Therefore, we can use the following equation:
h = v0y * t - (1/2) * g * t^2
Where h is the vertical distance, v0y is the initial vertical velocity, t is the time taken, and g is the acceleration due to gravity.
In this case, h = 6.5 m, t = 2.2 s, and g ≈ 9.8 m/s^2. Rearranging the equation, we get:
v0y = (h + (1/2) * g * t^2) / t
Substituting the values, we have:
v0y = (6.5 + (1/2) * 9.8 * 2.2^2) / 2.2
Calculate this and you will find the value of v0y, which represents the vertical component of the initial velocity.
Next, to find the horizontal component of the initial velocity, we can use the formula:
v0x = d / t
Where d is the horizontal distance and t is the time taken.
In this case, d = 24.0 m and t = 2.2 s. Substituting the values, we have:
v0x = 24.0 / 2.2
Calculate this and you will find the value of v0x, which represents the horizontal component of the initial velocity.
Finally, to find the speed at which the ball was launched, we can use the Pythagorean theorem:
v0 = √(v0x^2 + v0y^2)
Calculate this and you will get the speed at which the ball was launched.
b) To find the vertical distance by which the ball clears the wall, we need to consider the vertical motion of the ball. We already know the initial vertical velocity (v0y) and the time taken (t). The equation for vertical displacement is:
Δy = v0y * t - (1/2) * g * t^2
Substituting the values, we have:
Δy = v0y * 2.2 - (1/2) * 9.8 * 2.2^2
Calculate this and you will find the vertical distance by which the ball clears the wall.
c) To find the horizontal distance from the wall to the point on the roof where the ball lands, we can use the horizontal motion of the ball. The equation for horizontal distance is:
Δx = v0x * t
Substituting the values, we have:
Δx = v0x * 2.2
Calculate this and you will find the horizontal distance from the wall to the point on the roof where the ball lands.
By following these steps and performing the necessary calculations, you will be able to find the answers to all three questions.