Question

A ball is dropped onto a step at point A and rebounds with a velocity vo at an angle of 15° with the vertical. Determine the value of vo knowing that just before the ball bounces at point B its velocity vB forms an angle of 12° with the vertical. Also determine the velocity of B, vB and the distance d.

Answer 1

v₀ = 2.68 m/s.

`v_(B)` = 3.33 m/s

d = 1.61 m

Step-by-step explanation:

Searching on the internet I found the picture of the answer in which the height between point A and point B is equal to 0.2 m.

To determine the value of v₀ we need to use the following equation:

`v_(f) = v_(0) - gt`

Taking the initial vertical velocity (
`v_{0_(y)}`) as
`v_{A_(y)}` and the final vertical velocity (
`v_{f_(y)}`) as
`v_{B_(y)}` we have:

`v_{B_(y)} = v_{A_(y)} - gt`

We can express the time (t) in terms of the velocities of A and B:

`t = \frac{v_{A_(y)} - v_{B_(y)}}{g}` (1)

Now, we can use the equation:

`y_(f) = y_(0) + v_{A_(y)}t - (1)/(2)gt^(2)` (2)

By entering equation (1) into equation (2):

`y_(f) = y_(0) + v_{A_(y)}*(\frac{v_{A_(y)} - v_{B_(y)}}{g}) - (1)/(2)g(\frac{v_{A_(y)} - v_{B_(y)}}{g})^(2)` (3)

Since
`v_{A_(x)} = v_{B_(x)}` because there is no acceleration in the horizontal movement, we have:

`v_(A)sin(15) = v_(B)sin(12)`

`v_(A) = (v_(B)sin(12))/(sin(15))` (4)

Taking:

`y_(f)` = 0

`y_(0)` = 0.2 m

g = 9.81 m/s²

`v_{A_(y)} = v_(A)cos(15)`

`v_{B_(y)} = v_(B)cos(12)`

Entering the above values and equation (4) into equation (3) we have:

`-0.4 m*9.81 m/s^(2) = ((v_(B)sin(12)cos(15))/(sin(15)))^(2) - (v_(B)cos(12))^(2)`

By solving the above quadratic equation:

`v_(B) = 3.33 m/s`

Hence, the velocity of the ball
`v_(B)` is 3.33 m/s.

Now, we can find
`v_(A)` by using equation (4):

`v_(A) = (v_(B)sin(12))/(sin(15)) = (3.33 m/s*sin(12))/(sin(15)) = 2.68 m/s`

Then, the value of v₀ of the ball is 2.68 m/s.

Finally, to find the distance between point A and point B we need to calculate the time by using equation (1):

`t = \frac{v_{A_(y)} - (-v_{B_(y)})}{g} = \frac{v_{A_(y)} + v_{B_(y)}}{g}`

The minus sign in
`v_{B_(y)}` is because the vertical component of the vector is negative.

`t = \frac{v_{A_(y)} + v_{B_(y)}}{g} = (2.68 m/s*cos(15) + 3.33 m/s*cos(12))/(9.81 m/s^(2)) = 0.60 s`

Now, the distance is:

`x = v_(0)*t = 2.68 m/s*0.60 s = 1.61 m`

I hope it helps you!