Question

A projectile is fired at an upward angle of 55° from the top of a 120 m cliff with a speed of 150 m/s. What will be its speed when it strikes the ground below? (USE CONSERVATION OF ENERGY and neglect air resistance.) Express your answer to three significant figures and include the appropriate units.

Answer 1

Answer:

The speed when it strikes the ground below is V= 157.64 m/s < -56.92º .

Explanation:

V= 150m/s

α= 55º

hi= 120m

Vy= V*sinα

Vy= 122.87 m/s

Vx= V * cos α

Vx= 86.03 m/s

h= hi + Vy * t - g*t²/2

clearing t we get the total flying time of the projectile:

t(total fly)= 26.01 sec

0= Vy - g*t

clearing t we get the maximum height time:

t(max height)= 12.53 sec

to get the fall time:

t(fall)= t(total fly) - t(max height)

t(fall)= 13.48 sec

Vy'= g* t(fall)

Vy'= 132.1 m/s

V'= √(Vx² +Vy'²)

V'= 157.64 m/s

α'= tg⁻¹ (Vy'/Vx)

α'= -56.92º