1)A ball is thrown into air with initial velocity of 30m/s, 30 degree above horizontal. Determine :
A. Total time it stays in air
B. It is horizontal range.
C. Velocity of the ball 1.5 seconds later.
2)A football player attempts a 40 yard (36.58m) field goal. The goal post bar is 10 feet( 3.048m) above the field. If the ball is kicked from the ground with its initial x and y components of velocity equal to 15m/s, what is its height when it reaches the crossbar?
3) A ski jumper takes off from the ramp with an initial height of 40m, a horizontal velocity of 32m/s, and a vertucal velocity of 11m/s, landing with a final height of 1.5m. what is the direction of the takeoff velocity vector, the vertical velocity at landing, and the horizontal range?
4) A water balloon is launched with a velocity of 40m/s at 30 degree with an initial height of 0.8m. The balloon smashes into a building 15m away from the launch point. Find the following information at the moment of impact with the building :
A. horizontal and vertical velocity
B. height
C. time from launch to impact
D. direction of balloon's velocity vector.
Can you explain which formula to find the answer and how to use the correct way. Thankyou
physics motion,acceleration!
2011-03-08 7:42 pm
回答 (2)
2011-03-08 11:00 pm
✔ 最佳答案
1. (A) Consider the vertical velocity components,use equation: s = ut + (1/2)at^2with u = 30.sin(30) m/s = 15 m/s, a = -g(=-10 m/s2), s = 0 m, t = ?hence, 0 = 15t + (1/2).(-10)t^2i.e. t = 3 s (B) Horizontal range = 30.cos(30) x 3 m = 25.98 x 3 m = 77.94 m (C) use v = u + at for the vertical componentv = 15 + (-10).(1.5) m/s = 0 m/sHence, the ball is at the highest point. Velocity of the ball = 25.98 m/s, with direction horizontal. 2. Time for the ball to reach the crossbar = 36.58/15 s = 2.44 suse equation: s = ut + (1/2)at62 for the vertical componentss = 15 x 2.44 + (1/2).(-10).(2.44)^2 m = 6.83 m 3. Direction of take-off velocity = arc-tan[11/32] degrees above horizontal = 18.97 degreesUse equation: v^2 = u^2 + 2asv^2 = 11^2 + 2.(-10).(1.5-40)v = -29.83 m/sHence, vertical velocity at landing = 29.83 m/sTime of ski jumper in air = (-29.83-11)/(-10) s = 4.083 sHorizontal range = 32 x 4.083 m = 131 m 4. Horizontal velocity component = 40.cos(30) m/s = 34.64 m/sVertical velocity component = 40.sin(30) m/s = 20 m/sTime of flight = 15/34.64 s = 0.433 s(A) use equation: v = u + at for the vertical componentv = 20 + (-10).(0.433) m/s = 15.67 m/sThus, vertical velocity = 15.67 m/s, horizontal velocity = 34.64 m/s(B) use equation: s = (1/2)(u+v)ts = (1/2).(20+15.67).(0.433) m = 7.72 m (C) Time to impact = time of flight = 0.43 s(D) Direction of velocity = arc-tan[15.67/34.64] = 24.3 degrees above horizontal
2011-03-08 10:02 pm
收錄日期: 2021-04-29 17:40:35
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110308000051KK00446