Question

A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 3.80 m/s. If the roof is pitched at 38.0° below the horizon and the roof edge is 3.30 m above the ground, find the time the baseball spends in the air and the horizontal distance from the roof edge to the point where the baseball lands on the ground. HINT

Answer 1

a) The baseball spends 0.674 seconds in the air

b) The horizontal distance from the roof edge to the point where the baseball lands on the ground = 2.02 m

Step-by-step explanation:

The ball's initial speed, u = 3.8 m/s

θ = 38°

The edge of the roof has a height, H = 3.30 m

The vertical motion of the baseball can be given by the equation:

`H = U_(y) t + 0.5a_(y) t^(2)`.........(1)

Vertical acceleration of the baseball,
`a_y = 9.8 m/s^2`

The vertical component of the initial speed can be calculated by:

`U_y = Usin \theta\\U_y = 3.8 sin 38\\U_y = 2.34 m/s`

Substituting the appropriate values into equation (1):

`3.8 = 2.34 t + 0.5(9.8)} t^(2)\\4.9t^2 + 2.34t - 3.8 = 0`

Solving for t in the quadratic equation above:

t = 0.674 s

To calculate the horizontal distance, S, use the formula below:

`S = U_xt + 0.5a_xt^2`

Horizontal acceleration of the baseball,
`a_x = 0 m/s^2`

The horizontal component of the initial speed can be calculated as:

`U_x = Ucos \theta\\U_x = 3.8 cos 38\\ U_x = 2.99 m/s`

S = 2.99(0.674) + 0.5(0)

S = 2.02 m