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:
[tex]d=v_x t[/tex]
where:
[tex]v_x = 50 m/s[/tex] 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:
[tex]t=\frac{d}{v_x}=\frac{165}{50}=3.3 s[/tex]
Now we analyze the vertical motion: it is an accelerated motion, we can use the suvat equation
[tex]s=ut+\frac{1}{2}at^2[/tex]
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
[tex]a=g=9.8 m/s^2[/tex] is the acceleration of gravity
Solving for s, we find the height of the building:
[tex]s=0+\frac{1}{2}(9.8)(3.3)^2=53.4 m[/tex]
Learn more about projectile motion here:
brainly.com/question/8751410
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