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?

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

**Please read this before you use the other answer here!**

The angle in this problem is actually irrelevant. All you need is the INITIAL total energy and the FINAL total energy. So due to conservation of energy,

Total Energy (Initial) = Total Energy (Final)

Since the projectile has an initial velocity, AND it is starting above ground level, the initial Energy has both POTENTIAL and KINETIC. So, the initial Total Energy is

E_(i) = PE + KE = mgh + (1/2)mv_(i)²

When the projectile lands on the ground, the height is zero, so there is ONLY Kinetic Energy. So the final Total Energy is

E_(f) = (1/2)mv_(f)²

When these are set equal to each other, you have:

E_(i) = E_(f)

mgh + (1/2)mv_(i)² = (1/2)mv_(f)²

Then we solve for the final velocity: v_(f)

~ m cancels out ~

gh + (1/2)v_(i)² = (1/2)v_(f)²

~ multiply 2 to both sides ~

2gh + v_(i)² = v_(f)²

sqrt [2gh + v_(i)²] = v_(f)

Now plug in your values for g, h, and v_(i), and calculate the final velocity: v_(f).

(Note: This just calculates the speed, because we did not have to determine the direction / angle which would give us velocity.)

**(I mean no offense to the other tutor, but I believe this is the more accurate solution.)**