Question

A bowling ball is launched from the top of a building at an angle of 10° above the horizontal with an initial speed of 16 m/s. The ball lands on the ground 2.5 s after it is launched. What is the height of the building (in m)?

Answer 1

Answer:

height of the building, h = 23.68 m

Given:

angle, $\theta = 10^{\circ}$

initial speed, u = 16 m/s

time, t = 2.5 s

Solution:

To calculate the height of the building, we use second eqn of motion where, in falling down acceleration is acceleration due to gravity acting in the downward direction and vertical component of velocity usin$\theta$ is considered and the distance travelled is the height of the body:

$h = usin\theta t - \frac{1}{2}gt^{2}$                    

Putting the values in the above eqn:

$h = 16sin10^{\circ}\times 2.5 - \frac{1}{2}\times 9.8\times (2.5)^{2}$    

h =  23.68 m            

Answer 2

The height of the building is 37.6 m

The given parameters;

• angle of projection, Ф = 10⁰

• initial velocity of the projectile, v = 16 m/s

• time of motion of the ball from the top building, t = 2.5 s

The height of the building form the which the ball was projected is calculated as shown below;

$h = v_0_y t+ \frac{1}{2} gt^2\\\\h = (v\times sin(\theta) )\times t \ + \ \frac{1}{2} gt^2\\\\h = (16 \times sin(10) )\times 2.5 \ + (0.5 \times 9.8 \times 2.5^2)\\\\ h = 37.6 \ m$

Thus, the height of the building form the which the ball was projected is 37.6 m

Learn more here: https://brainly.com/question/15162556