Question

A ball is kicked from the top of a building with a velocity of 50 m/s and lands 165 m away from the base of the building.

Answer 1

The height of the building is 53.4 m (projectile motion)

Explanation:

The motion of the ball is a projectile motion, so it follows a parabolic path which consists of two independent motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction  

First of all, we consider the horizontal motion, which is a uniform motion at constant speed. The equation that gives the horizontal distance travelled is:

$d=v_x t$  

where:

$v_x = 50 m/s$ is the horizontal velocity of the ball, which is constant since there are no forces in this direction

d = 165 m is the horizontal distance travelled by the ball

Solving for t, we find the time of flight:

$t=\frac{d}{v_x}=\frac{165}{50}=3.3 s$

Now we analyze the vertical motion: it is an accelerated motion, we can use the suvat equation

$s=ut+\frac{1}{2}at^2$  

where  

s is the vertical displacement, which is the height of the building

u = 0 is the initial vertical velocity of the ball

t = 3.3 s is the time of flight

$a=g=9.8 m/s^2$ is the acceleration of gravity  

Solving for s, we find the height of the building:

$s=0+\frac{1}{2}(9.8)(3.3)^2=53.4 m$

Learn more about projectile motion here:

brainly.com/question/8751410

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