A rock is dropped from the top of a 45m tower and at the same time a ball is thrown from the top of the tower in a horizontal direction. Air resistance is negligible. The ball and the rock hit the level ground a distance of 30m apart. The horizontal velocity of the ball thrown was almost nearly....?
I know the answer is 10 m/s but why? Can someone explain how you get the answer?
thanks
2006-11-05 15:30:29 · 3 answers · asked by odonata 2 in Science & Mathematics ➔ Physics
Don't even worry about the rock yet. The ball takes a certain time to drop 45m vertically and in that time it goes 30m horizontally. Air resistance is negligible so the horizontal velocity is constant. In the vertical direction, the ball starts motionless and accelerates and 9.81 m/s^2.
Time to fall 45m
45 = (1/2)(9.81)t^2
t = 3.0289s
horizontal velocity:
30 = 3.0289*v
v = 9.9m/s
2006-11-05 15:40:42 · answer #1 · answered by cliffrosewood 1 · 0⤊ 0⤋
Using the rock will give you the time until both objects hit bottom using at^2/2 = d , a = 10 and d = 45 gives t of 3 seconds. So the horizontal distance gives d = vt with d = 30m and t = 3 to give a v of 10m/sec
The horizontal and vertical motion can be treated independently.
2006-11-05 15:37:36 · answer #2 · answered by rscanner 6 · 0⤊ 0⤋
The time to fall is t = sqrt(2h/a) = sqrt(2*45/9.8) = 3.03 sec
If the ball has a consant horizontal velocity Vh, it must cover the 30 m in 3.03 sec, so
Vh = 30/3.03 = 9.9 m/s........Within 1% of your answer!
2006-11-05 15:38:51 · answer #3 · answered by Steve 7 · 0⤊ 0⤋