A ball rolls off a table 1.45m high. The ball lands 1.21m from the base of the table.
This can be solved with x / y components.
a) How much time does the ball take to hit the floor?
b) How fast was the ball going when it rolled off the table?
c) What is the velocity with which the ball hits the floor? (not zero) With magnitude and direction.
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Consider the Y axis in the vertical direction and the X axis in the horizontal direction. In the vertical direction, the ball moves with a uniformly varied movement, and in the horizontal direction makes a uniform motion.
First we calculate the time it takes the ball to reach the floor; For this we use the following equation:
h = v0yt + (gt^2)/2
h, is the height of the table.
v0y, is the initial velocity of the ball in the vertical direction.
g, is the acceleration of gravity.
t, is the time of fall.
In this case the output speed of the ball at the end of the table is directed in horizontal direction, so that the initial velocity in the vertical direction is zero. Solving the equation for the time we have:
h = (gt^2)/2
t = sqrt (2h/g)
t = sqrt (2*1.45/9.8) = 0.54 s
Since in the horizontal direction the ball travels 1.21 m, with uniform motion, in the same time we found in part (a), we can apply the following equation, to calculate the output speed (vx) from the table:
vx = d/t = (1.21)/(0.54) = 2.24 m/s
To find the speed with which the ball hits the ground, we calculate the final vertical speed using the following equation:
vy = v0y + gt
vy = gt = (9.8)(0.54) = 5.29 m/s
Then the speed with which the ball hits the floor is:
v = sqrt (vx^2 + vy^2)
v = sqrt (5.01 + 27.98)
v = 5.74 m/s
To find the direction of this speed, we use the tangent of the angle relative to the horizontal direction:
tag θ = vy/vx = (5.29)/(2.24) = 2.36
The direction of the velocity is:
θ = 67.05°
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