A projectile is fired straight up from the top of a 190m cliff with a speed of 170m/s . What will be its speed just before it strikes the ground below? (Use conservation of energy and neglect air resistance.)
when will it hit?
190 + 170t - 4.9t^2 = 0
t = 35.777
what is its speed?
v = 170-9.8t = 170-9.8*35.77 = -180.546
Or, using conservation of energy,
max height is reached when v=0, so
170-9.8t=0
t=17.346
h(17.346) = 1664.49
At that point, the
PE = mgh = 1664.49*9.8m = 16312m
So, final KE must be the same:
1/2 mv^2 = 16312m
v^2 = 32624
v = 180.62
Close enough for jazz.