A daredevil jumps a canyon 14 m wide. To do so, he drives a car up a 13 degree incline.
The acceleration of gravity is 9.81 m/seconds squared
a) What minimum speed must he achieve to clear the canyon? Answer in units of m/s
b0 If the daredevil jumps at this minimum speed, what will his speed be when he reaches the other side? Answer in units of m/s
2006-09-17 17:25:23 · 2 answers · asked by cantthinkofgoodname2001 1 in Education & Reference ➔ Homework Help
He must have enough vertical velocity vy so that the time it takes to reach apex and fall back is equal to the time it takes for the horizontal velocity to span the canyon. vy-at gives the time to reach apex; his total time aloft is twice this. t=2*vy/a; the time to cross the canyon is given by w = vx*t, or t = w/vx. so equate these to get
2*vy/a = w/vx.
The takeoff angle gives the relation between vy and vx also: vy/vx = tan(takeoff angle). Now you can solve for vx and vy. The takeoff speed is the vector sum of these, sqrt(vx^2+vy^2). His landing speed will be the same, as the situation is symmetric: vx doesn't change, and vy' will be at, where t = vy/a so vy' = vy
2006-09-17 17:42:55 · answer #1 · answered by gp4rts 7 · 0⤊ 0⤋
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2006-09-18 01:45:07 · answer #2 · answered by Sham 2 · 0⤊ 0⤋