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?

Distance, y = -2.287 meters

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,

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

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

t is the time of flight

t = 0.828 seconds

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

Here, a = -g

is the vertical component of velocity

y = -2.287 meters

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