A ball rolls down an incline that ends in a drop-off. The incline makes a constant angle of 20 degrees below the horizontal. The balls rolls down this incline for 30 m, accelerating at a constant 4 m/s^2. At this point it reaches the edge of the drop-off a falls a distance of 5 m. At what horizontal distance from the edge of the drop-off does the ball hit the ground?
The answer my teacher gave was 8.8 m and I am just not getting that! Thank you in advance!
The answer my teacher gave was 8.8 m and I am just not getting that! Thank you in advance!
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find the velocity of the ball as it leaves the ramp.
v^2 = 2*a*s => v = √(2*4*30) = 15.5 m/s (30° below horizontal)
to find the horizontal distance, you first have to calculate the time it takes to fall:
y = y₀ + v₀t + (1/2)*a*(t^2)
In this equation, we'll solve for t by using the velocity and acceleration in y-direction:
5 = (15.5 sin 20)t + (1/2)*(9.8)*(t^2)
t = .61 s
now find the horizontal distance with the same kinematic equation, but this time in the x-direction:
x = x₀ + v₀t + (1/2)*a*(t^2) ; (but acceleration in the x-dir = 0): Δx = v₀t
Δx = (15.5 cos 20)*(.61)
this indeed comes out to be Δx = 8.8m
v^2 = 2*a*s => v = √(2*4*30) = 15.5 m/s (30° below horizontal)
to find the horizontal distance, you first have to calculate the time it takes to fall:
y = y₀ + v₀t + (1/2)*a*(t^2)
In this equation, we'll solve for t by using the velocity and acceleration in y-direction:
5 = (15.5 sin 20)t + (1/2)*(9.8)*(t^2)
t = .61 s
now find the horizontal distance with the same kinematic equation, but this time in the x-direction:
x = x₀ + v₀t + (1/2)*a*(t^2) ; (but acceleration in the x-dir = 0): Δx = v₀t
Δx = (15.5 cos 20)*(.61)
this indeed comes out to be Δx = 8.8m