AP Physics due 2nite please help????? anyone please im begging you??

Question

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

Answer 1

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.

Answer 2

You can do part b before part a - it makes it simpler.

Part b -> It takes the dog as much time to go up as come down and at the top of his trajectory he has 0 m/s speed in the vertical direction. So use:

v = 0 = v0*sin(q) -g*t --> t = v0*sin(q)/g where q = angle of takeoff

Now the dog is in the air for 2*t = 1.1 seconds

He travels a distance along the ground given by

x = v0*cos(q)*2*t = 12.61 m

Answer 3

6 feet ,3 seconds