Question

A soccer ball is kicked from the top of one building with a height of H1 = 32.0 m to another building with a height of H2 = 13.8 m. (It is not a very smart idea to play soccer on the roof of tall buildings.)The ball is kicked with a speed of v0 = 16.1 m/s at angle of ? = 78.0° with respect to the horizontal. How much time will the ball spend in the air before it hits the roof of the other building? (The soccer ball is kicked without a spin. Neglect air resistance.)

Answer 1

Answer:

2.3s

Explanation:

The initial velocity in the vertical direction:

$v_v = v_0sin(\alpha) = 16.1sin(78^0) = 16.1*0.208 = 3.35 m/s$

Since the ball is kicked from H1 = 32m to another building at H2 = 13.8m. Vertically speaking it has traveled a distance of 13.8 - 32 = -18.2 m with respect to the initial building

We can use the following equation of motion to solve for the time t it takes to travel

$s = v_vt + gt^2/2$

where s = -18.2 m is the distance traveled with respect to the first building, g = -9.81 m/s2 is the gravitational acceleration in the opposite direction with the initial velocity

$-18.2 = 3.35t -9.81 t^2/2$

$4.905t^2 - 3.35t - 18.2 = 0$

$t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$t= \frac{3.35\pm \sqrt{(-3.35)^2 - 4*(4.905)*(-18.2)}}{2*(4.905)}$

$t= \frac{3.35\pm19.19}{9.81}$

t = 2.3 or t = -1.61

Since t can only be positive we will pick t = 2.3s