A parachutist with a camera descends in free fall at a speed of 16 m/s. The parachutist releases the camera at an altitude of 70 m.
(a) How long does it take the camera to reach the ground?
b) What is the velocity of the camera just before it hits the ground?
i got 5.75 s and 44.2 m/s.... and got both wrong.... please help
2007-10-11 11:14:14 · 2 answers · asked by Pauly Walnuts 2 in Science & Mathematics ➔ Physics
Use s = ut + 1/2 a t^2 to find t.
s = 70, u = 16, and a = 9.8, so
70 = 16t + 1/2 x 9.8 x t^2, so
70 = 16t + 4.9t^2, so
4.9t^2 + 16t - 70 = 0.
use the quadratic formula to solve for t, and then use your value of t in v = u + at to find the solution for part b.
Hope this helps, Twiggy.
2007-10-11 11:27:46 · answer #1 · answered by Twiggy 7 · 0⤊ 0⤋
D = ViT + .5AT^2
70 = 16T + .5(9.8)T^2
solve this quadratic equation for T
Vf = Vi + AT
Vf = 16 +9.8T
put in the time you got from above.
This give you the final velocity
2007-10-11 18:30:56 · answer #2 · answered by Jeffrey K 7 · 0⤊ 0⤋