Caitlin is playing tennis against a wall. She hits the tennis ball from a height of 0.5 m above the ground with a velocity of 20.0 m/s at an angle of 15.0° to the horizontal toward the wall that is 6.00 m away. a) How far off the ground is the ball when it hits the wall? b) Is the ball still traveling up or is it on its way down when it hits the wall?
2007-10-22 15:25:23 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Anyone? I just can't seem to figure it out
2007-10-23 00:44:32 · update #1
The ball will reach the wall in 0.3106 seconds.
6/[20*cos(15)]=0.3106
The vertical component of the ball's initial velocity is 5.1764 m/s.
20*sin(15)=5.1764
The vertical component will be reduced by 9.8 m/s/s.
At the end of 0.3106 seconds the vertical component will still be 2.1327 m/s upward.
5.1764-(9.8*0.3106)=2.1327
The ball hits the wall at a height of 1.5802 meters.
6*sin(15)+.5-(0.3106^2*4.9)=1.5802
2007-10-30 10:45:38 · answer #1 · answered by farwallronny 6 · 0⤊ 0⤋
First, compute v0[x], v0[y] = cos,sin(theta)*v0 = 19.3185 m/s, 5.1764 m/s
Then flight time t = range/v0[x] = 0.3106 s
A. y = y0 + v0[y]*t-gt^2/2 = 0.5 m + 1.1350 m = 1.635 m
B. vy = v0[y]-gt = 2.1327 m/s
Since vy is still positive it's travelling up.
2007-10-27 09:34:47 · answer #2 · answered by kirchwey 7 · 0⤊ 0⤋