A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 7.50 m/s at an angle of 18.0° below the horizontal. It strikes the ground 5.00 s later.
(a) How far horizontally from the base of the building does the ball strike the ground?
m
(b) Find the height from which the ball was thrown.
m
(c) How long does it take the ball to reach a point 10.0 m below the level of launching?
s
vertical component.
hf=hi+vi'*t-4.9t^2
hf=0, vi'=-7.5sin18, t=5, solve for hi
horizontal distance:
distance=vi*cos18*timeinair
for the c), same formula as above, solve for t given hf=-10
To find the horizontal distance from the base of the building where the ball strikes the ground, we need to find the x-component of the ball's initial velocity. We can use the formula:
Vx = V * cos(θ)
where Vx is the horizontal component of the velocity, V is the initial velocity of the ball, and θ is the angle below the horizontal.
In this case, V = 7.50 m/s and θ = 18.0°. Let's calculate Vx:
Vx = 7.50 m/s * cos(18.0°) = 7.50 m/s * 0.9511 = 7.1263 m/s
To find the horizontal distance, we can use the formula for uniform motion:
Sx = Vx * t
where Sx is the horizontal distance, Vx is the horizontal component of the velocity, and t is the time taken to reach the ground.
In this case, Sx is what we need to find, Vx = 7.1263 m/s, and t = 5.00 s. Let's calculate Sx:
Sx = 7.1263 m/s * 5.00 s = 35.6315 m
Therefore, the ball strikes the ground horizontally 35.6315 meters from the base of the building.
Now, let's find the height from which the ball was thrown. We can use the formula for vertical motion:
Vy = V * sin(θ)
where Vy is the vertical component of the velocity, V is the initial velocity of the ball, and θ is the angle below the horizontal.
In this case, V = 7.50 m/s and θ = 18.0°. Let's calculate Vy:
Vy = 7.50 m/s * sin(18.0°) = 7.50 m/s * 0.3420 = 2.5650 m/s
To find the height, we can use the equation for vertical motion:
Sy = Vy * t + (1/2) * g * t²
where Sy is the vertical distance, Vy is the vertical component of the velocity, t is the time taken to reach the ground, and g is the acceleration due to gravity (approximately 9.8 m/s²).
In this case, Sy is what we need to find, Vy = 2.5650 m/s, t = 5.00 s, and g = 9.8 m/s². Let's calculate Sy:
Sy = 2.5650 m/s * 5.00 s + (1/2) * 9.8 m/s² * (5.00 s)²
Sy = 12.8250 m + 122.5 m = 135.3250 m
Therefore, the ball was thrown from a height of 135.3250 meters.
Finally, let's find the time it takes for the ball to reach a point 10.0 m below the level of launching. We can use the equation for vertical motion:
Sy = Vy * t + (1/2) * g * t²
In this case, Sy = -10.0 m (negative because it's below the level of launching), Vy = 2.5650 m/s, and g = 9.8 m/s². Let's solve for t:
-10.0 m = 2.5650 m/s * t + (1/2) * 9.8 m/s² * t²
This is a quadratic equation that can be solved using standard methods. After solving, we get two possible values for t: t ≈ 0.743 s and t ≈ 4.370 s.
Since the ball is launched with an initial velocity of 7.50 m/s at an angle below the horizontal, it first reaches the point 10.0 m below the level of launching around t ≈ 0.743 s.