Question

A superhero launches himself from the top of a building with a velocity of 7.3m/s at an angle of 25 above the horizontal. If the building is 17 m high, how far will he travel horizontally before reaching the ground? What is his final velocity?

Answer 1

Given info : A superhero launches himself from the top of a building with a velocity of 7.3m/s at an angle of 25 above the horizontal. if the building is 17m high.

To find : how far will he travel horizontally before teacher the ground ? and what is his final velocity ?

solution : initial velocity, u = 7.3 m/s

angle of inclination, θ = 25°

now, $u_y$ = 7.3sin25° = 3.085 m/s

using formula, $y=u_yt+\frac{1}{2}a_yt^2$

⇒-17 = 3.085 × T + 1/2 × -9.8 × T²

⇒ 4.9T² - 3.085T - 17 = 0

⇒T = -1.5 , 2.2 but time can't be negative

so, T = 2.2 sec

Therefore the time of flight is 2.2 sec.

now horizontal initial velocity, $u_x$ 7.3cos25° = 6.616 m/s

and horizontal distance travelled by superhero = $u_xT$

= 6.616 × 2.2 m

= 14.56 m

and final velocity, $v=u_x\hat{i}+(u_y-gT)\hat{j}$

= 7.3cos25° i + (7.3 sin25° - 9.8 × 2.2)j

= 6.616 i + (3.085 - 21.56)j

= 6.616i - 18.475j

magnitude of final velocity , v = 19.62 m/s