Question

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?

Answer 1

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.