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 22.0° below the horizon and the roof edge is 4.90 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.

Answer 1

Answer:

Time = 0.86 s

Horizontal distance = 3.03 m

Explanation:

Given data:

initial velocity = $v_{i}$ = 3.8 m/s

θ = 22° below with horizontal

Height = h = 4.9 m

a.) Time = t = ?

b.) Horizontal distance = R = ?

a.) First we need to find time of flight

Resolve Velocity into horizontal and vertical component

Horizontal component = $v_{i}_{x}$ = $v_{i}$Cosθ

                                         = 3.8Cos22°

                       $v_{i}_{x}$     = 3.52 m/s

Vertical component = $v_{i}_{y}$= $v_{i}$Sinθ

                                     = 3.8Sin22°

                           $v_{i}_{y}$   = 1.42 m/s

Using 2nd equation of motion

            $h = v_{i}_{y}t+\frac{1}{2}gt^{2}$

                 4.9 = 1.42t + 4.9t²

the above equation is quadratic. So it has 2 outputs. By solving above equation we have two outputs that is

                 t = 0.86 s            &               t  = -1.15 s  

Time can never be negative ,So the correct answer is t = 0.86 s

                                  t = 0.86s

b.)    

   As the horizontal component of velocity in projectile motion remain constant, so there is no acceleration along horizontal.

we can simply use this formula

                 R = $v_{i}_{x}t$

                  R = (3.52)(0.86)

                  R = 3.03 m