A projectile is fired at an upward angle of 45.0 ∘ from the top of a 170 m cliff with a speed of 180 m/s .

Question

A projectile is fired at an upward angle of 45.0 ∘ from the top of a 170 m cliff with a speed of 180 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

Our intuition would tell us to just use the equations that relate velocity, acceleration, and time. However, all those equations require one key value, the mass of the projectile, which we don't have!

This is why the problem statement calls for the use of the conservation of energy, so let's write it out:

1. PE₁ + KE₁ = PE₂ + KE₂

1. m*g*h + (1/2) m*(v₁)^2 = 0 +(1/2) m*(v₂)^2

Fortunately, the mass terms cancel out! We can then simplify the 2^nd equation to:

1. v₂ = √(2*g*h+v₁²)

v₂ = √(2*(9.81 m/s^2)*(170 m)+(180 m/s)²)

v₂ = 189 m/s