Question

A guy was dumped by his girlfriend and wants to jump off from a building 200 ft high. What is the required velocity he must start in order to touch down on her new boyfriend who stands on the ground, 30ft from the building. a) if he jumped out horizontally b)the guy jumped down with a depression angle of 30 degrees. Find the impact velocity and speed. Does he have any chance?

Answer 1

a)

consider the motion in vertical direction :

V_oy = initial velocity = 0 m/s

a_y = acceleration = - 32.2 m/s²

Y = vertical displacement = - 200 ft

t = time of travel = ?

using the equation

Y = V_oy t + (0.5) a t²

- 200 = 0 t + (0.5) (- 32.2) t²

t = 3.5 sec

consider the motion in horizontal direction :

V_ox = horizontal velocity = ?

X = horizontal distance to be travelled = 30 ft

Using the equation

V_ox = X/t = 30/3.5 = 8.6 ft/s

consider the motion along X-direction :

V_fx = final velocity at the time of impact = V_ox = 8.6 ft/s

since there is no acceleration along x-direction

Consider the motion along Y-direction

V_fy = V_oy + a_y t

V_fy = 0 + (- 9.8) (3.5) = 34.4 m/s

Net velocity is given as

V_f = sqrt(V_fx² + V_fy²) = sqrt((8.6)² + (34.4)²) = 35.5 m/s

b)

consider the motion in horizontal direction :

V_ox = horizontal velocity = v Cos30

X = horizontal distance to be travelled = 30 ft

Using the equation

V_ox = X/t

v Cos30 = 30/t

t = 30/(v Cos30) eq-1

consider the motion in vertical direction :

V_oy = initial velocity = -v Sin30 m/s

a_y = acceleration = - 32.2 m/s²

Y = vertical displacement = - 200 ft

t = time of travel = ?

using the equation

Y = V_oy t + (0.5) a t²

- 200 = (v Sin30) (30/(v Cos30)) + (0.5) (- 32.2) (30/(v Cos30))²

t = 9.43 sec

using eq-1

t = 30/(v Cos30)

9.43 = 30/(v Cos30)

v = 3.7 m/s

consider the motion along X-direction :

V_fx = final velocity at the time of impact = V_ox = v cos30 = 3.7 Cos30 = 3.2 ft/s

since there is no acceleration along x-direction

Consider the motion along Y-direction

V_fy = V_oy + a_y t

V_fy = (3.7 Sin30) + (- 9.8) (3.5) = - 32.5 m/s

Net velocity is given as

V_f = sqrt(V_fx² + V_fy²) = sqrt((3.2)² + (- 32.5)²) = 32.7 m/s