First of all the mass of the rock has nothing to do with this problem.
if it is thrown straight up at at Vi = 19 m/s from a height of 1.75 meters:
V = Vi - g t
at top, v = 0
0 = 19-9.81 t
t = 1.94 seconds to top
h = 1.75 + 19(1.94) -4.9(1.94)^2
h = 20.2 meters high
if it is thrown at an angle and has a speed of 8 at the top.
At the top, motion is horizontal.
So
u = horizontal speed = 8 m/s the whole time.
u = 8 = 19 cos A
where A is angle up from horizontal at launch
so
A = 65 degrees launch angle
Vi = 19 sin 65 = 17.2 initial speed up
v = 17.2 - gt
t = 1.76 seconds to max height
h = 1.75 + 17.2(1.76) - 4.9(1.76^2)
h = 16.8 meters
Bryan tosses a 75.0 g rock into the lake. the rock leaves hus hand at a height of 175 cm above the ground travelling at 19.0 m/s. How high above the ground will the rock travel if it us accidentally bryan tosses a 75.0 g rock into the lake. The rock leaves hus hand at a height of 175 cm above the ground travelling at 19.0 m/s.
How high above the ground will the rock travel if it is accidentally thrown straight up?
How high above the water will it travel if it is thrown in an arc and has a speed of 8.00 m/s at its maximum height?
1 answer
1 answer