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In a scene in an action movie, a stuntman jumps from the top of another building that is 4.0 m away. After a running start, he leaps at a velocity of 5.0 m/s at an angle of 15 degrees with respect to the flat roof. How far will he fall?

1 Answer

4 votes

Answer:

Distance, y = -2.287 meters

Step-by-step explanation:

Given that,

The separation between tow buildings, d = 4 m

Velocity of the stuntman, v = 5 m/s

Angle with respect to the flat roof,
\theta=15^(\circ)

We know that the horizontal component of velocity is given by :


v_x=v\ cos\theta

The velocity of an object is equal to the total displacement divided by total time taken as :


v=(d)/(t)

t is the time of flight


t=(d)/(v_x\ cos\theta)


t=(4)/(5\ cos(15))

t = 0.828 seconds

Let he will fall to d distance of y. Using the second equation to find it as :


y=vt+(1)/(2)at^2

Here, a = -g


y=u_yt-(1)/(2)gt^2


u_y is the vertical component of velocity


y=u\ sin\theta t-(1)/(2)gt^2


y=5\ sin(15)* 0.828-(1)/(2)* 9.8* (0.828)^2

y = -2.287 meters

So, he will fall at a distance of 2.287 meters. Hence, this is the required solution.

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