Question

Problem 6.38 A projectile is fired at an upward angle of 40.0 degree from the top of a 245-m cliff with a speed of 235m/s. Part A What will be its speed when it strikes the ground below? (Use conservation of energy.) V =m/s

Answer 1

Answer 2

Note that the angle of the projectile does not matter since the angle does not change the amount of energy that the bullet possesses.

The sum of the potential and kinetic energy at launch is:
E(initial) = PE + KE = mgh(cliff) + (1/2)mv(i)^2.

At the ground, the energy is all kinetic, so:
E(final) = KE = (1/2)mv(f)^2.

Since energy is conserved:
mgh(cliff) + (1/2)mv(i)^2 = (1/2)mv(f)^2
==> v(f) = ?[2gh(cliff) + v(i)^2]
= ?[2(9.8)(245) + 235^2]
= 232 m/s (to 2 s.f.).

Answer 3

conserving energy,

0.5*mv^2=mgh+0.5mv^2

or 0.5*v^2=9.81*235+0.5*245^2

or v=254.235 m/s