GIVEN: A cannonball shot with an initial velocity of 141 m/s at an angle of 36° follows a parabolic path and hits a balloon at the top of its trajectory.
____ m/s
2007-01-23 03:57:19 · 3 answers · asked by Khoi 1 in Science & Mathematics ➔ Physics
The initial velocity of 141 m/s is at an angle of 36°. That means that the initial velocity is in two directions, up and to the right. Some of that 141 m/s is to the right, some of it is up. The amount that is up is (141 m/s)(sin 36) = 82.9 m/s, the amount to the right is (141 m/s)(cos 36) = 114 m/s. When the cannonball is at the top of its trajectory, the vertical component of the velocity is 0. That is, the 82.9 m/s is gone, it has been reduced by gravity. The velocity to the right, 114 m/s, is unchanged. Thus, the final velocity at the top of its trajectory is 114 m/s.
2007-01-23 04:05:31 · answer #1 · answered by Nicknamr 3 · 2⤊ 0⤋
Derive initial velocity to X and Y axis
X axis velocity = 141 (cos 36)
Y axis = 141 (sin 36)
Top of trajectory means Y component reach zero
we only have 141 (cos 36) which is 114.07 m/s
2007-01-23 04:08:52 · answer #2 · answered by EggsarefromChicken 2 · 0⤊ 0⤋
141xcos(36)
2007-01-23 04:12:41 · answer #3 · answered by ag_iitkgp 7 · 0⤊ 0⤋