that depends on how fast the 2nd ball is launched.
If it is sent up at time t=a, then
ball 1.
y = 180-4.9t^2
x = 20t
ball 2:
y = v sin15° (t-a) - 4.9(t-a)^2
x = 20√85 + v cos15° (t-a)
Without knowing v, there is no way to get a final number for a.
A ball is thrown horizontally from the top of a tower @180m high with a speed of 20m/s how much later must another ball be projected at an angle of 15° to the horizontal from a point on the ground 20√85m away from the top of the tower if the two balls are to meet at the same point
1 answer
1 answer