Help with physics projectile question?

Question

A cannonball is fired from a cliff 140 m high downward at an angle of 27 degrees with respect to the horizontal. If the cannon's muzzle velocity is 34 m/s, what is the cannon ball's speed when it hits the ground?

I am not sure how to do this problem...I would appreciate any help on how to do it or example problems. (my book dosn't have any it just tells you the theory...)

Answer 1

There's a much easier solution than the one given by civil_av8r.

The angle of the cannon makes no difference, so ignore that.

Use the principle of conservation of energy.

At the top, the cannonball's total energy is:
PE_initial + KE_initial
= mgh + mv²/2
= mg(140meters) + m(34meters/sec)²/2

At the bottom, the cannonballs total energy is:
PE_final + KE_final
= 0 + m(v_final)²/2

By conservation of energy:
PE_initial + KE_initial = PE_final + KE_final
mg(140meters) + m(34meters/sec)²/2 = m(v_final)²/2

Divide by "m" to make it cancel out:
g(140meters) + (34meters/sec)²/2 = (v_final)²/2

Use algebra to solve for v_final:
v_final = sqrt(2g(140meters) + (34meters/sec)²)

Answer 2

I'll show you the method, you can plug in the numbers

vx = 34*cos(27) m/s
vy = -34*sin(27) m/s

1st thing is to figure out how long it takes to hit the ground

s = s0 + vy*t - 1/2gt^2
s0 = 140
s = 0

(-1/2*g)t^2 + (-34*sin(27))t + 140 = 0

solve for t using the quadratic equation

find the final velocity in the y direction when it hits the ground

vyf = vyi - g*t
vyf = -34*sin(27) - g*t

final velocity = sqrt(vyf^2 + vx^2)