When chasing a hare along a flat stretch of ground, a greyhound leaps into the air at a speed of 12.7 m/s, at an angle of 25° above the horizontal.
(a) What is the range of his leap?
____m
(b) For how much time is he in the air?
_____s
2007-09-17 15:05:22 · 1 answers · asked by Antonio E 1 in Science & Mathematics ➔ Physics
Well, if he leaps at a 25 degree angle at 12.7 m/s then his horizontal speed will be 12.7*cos (25 degrees) m/s and his vertical sped will be 12.7*sin (25 degrees) m/s. The force of gravity will accelerate the dog down at 9.8 m/s^2.
So the equation to see how long he stays in the air would be d = 12.7*sin (25 degrees) m/s * t - 9.8 m/s^2 * t^2. If we want to know the time when he hits to ground then we just set d to 0.
0 = 12.7sin (25 degrees) m/s * t - 9.8 m/s^2 * t^2
9.8 m/s^2 * t^2 = 12.7sin (25 degrees) m/s * t
9.8 m/s^2 * t = 12.7sin (25 degrees) m/s
t = 12.7sin (25 degrees) / 9.8 seconds
Which equals about .55 seconds.
To get the range of his leap is easy from here. It's how long he stayed in the air times his horizontal speed:
12.7 cos (25 degrees) * 12.7sin (25 degrees) / 9.8 seconds
Which equals about 6.30 meters
a) about 6.30 meters
b) about 0.55 seconds
Hope this helps.
2007-09-17 15:19:59 · answer #1 · answered by someone2841 3 · 0⤊ 0⤋