Question

1. -/1 points a bowling ball is launched from the top of a building at an angle of 20° above the horizontal with an initial speed of 15 m/s. the ball lands on the ground 3.2 s after it is launched. what is the height of the building (in m)? m resources read it

Explanation:

Step1: Analyze vertical - motion

The vertical - motion of the ball is a uniformly - accelerated motion. The initial vertical velocity is $v_{0y}=v_0\sin\theta$, where $v_0 = 15$ m/s and $\theta = 20^{\circ}$, and the acceleration $a=-g=- 9.8$ m/s², and the time of flight $t = 3.2$ s. The displacement in the vertical direction $y - y_0$ is related to the initial velocity, acceleration, and time by the equation $y - y_0=v_{0y}t+\frac{1}{2}at^{2}$. The height of the building $h$ is the magnitude of the vertical displacement when the ball lands on the ground ($y - y_0=-h$).
$v_{0y}=v_0\sin\theta=15\times\sin20^{\circ}\approx15\times0.342 = 5.13$ m/s.

Step2: Apply the kinematic equation

Using the kinematic equation $y - y_0=v_{0y}t+\frac{1}{2}at^{2}$, substituting $y - y_0=-h$, $v_{0y}=5.13$ m/s, $a=-9.8$ m/s², and $t = 3.2$ s.
$-h=v_{0y}t+\frac{1}{2}at^{2}=5.13\times3.2+\frac{1}{2}\times(-9.8)\times3.2^{2}$.
First, calculate $5.13\times3.2 = 16.416$ m.
Second, calculate $\frac{1}{2}\times(-9.8)\times3.2^{2}=-4.9\times10.24=-50.176$ m.
Then, $-h=16.416-50.176=-33.76$ m.
So, $h = 33.76$ m.

Answer:

$33.76$