Question

A projectile is fired at an upward angle of 380 degrees from the top of a 105-m-high cliff with a speed of 165m/s. What is its speed when it strikes the ground below ?

Answer 1

Angle of firing of projectile = = 38^o

Initial speed of the projectile = V₀ = 165 m/s

The projectile is fired from the top of a 105m high cliff and it strikes the ground in the end.

Displacement in the vertical direction = y = -105 m

Initial velocity in vertical direction = V_y0

V_y0 = V₀Sin

V_y0 = 165Sin(38)

V_y0 = 101.58 m/s

Initial velocity in horizontal direction = V_x0

V_x0 = V₀Cos

V_x0 = 165Cos(38)

V_x0 = 130.02 m/s

Time taken by the projectile to hit the ground = t

Speed of the projectile when it strikes the ground = V₁

Vertical velocity of the projectile when it strikes the ground = V_y1

Horizontal velocity of the projectile when it strikes the ground = V_x1

There is no force acting on the projectile in the horizontal direction, therefore the velocity in the horizontal direction remains constant.

V_x1 = V_x0

V_x1 = 130.02 m/s

Acceleration in the vertical direction = -g = -9.81 m/s²

In the vertical direction,

y = V_y0t + (-g)t²/2

-105 = (101.58)t - (9.81)(t²)/2

4.905t² - 101.58t - 105 = 0

t = 21.7 or -0.98 sec

Time cannot be negative

t = 21.7 sec

V_y1 = V_y0 + (-g)t

V_y1 = 101.58 - (9.81)(21.7)

V_y1 = -111.3 m/s

V₁ = 171.15 m/s

Speed of the projectile when it strikes the ground = 171.15 m/s