A robot probe drops a camera off the rim of
a 315 m high cliff on Mars, where the free-fall
acceleration is 3.7 m/s2
.
a) Find the velocity with which it hits the
ground.
Answer in units of m/s.
b)Find the time required for the camera to reach
the ground.
Answer in units of s.
Help with Physics question?
2016-06-08 5:24 am
回答 (3)
2016-06-08 6:13 am
a) v^2 = 2gh
v^2 = 2 x 3.7 x 315
v = 48.3 m/s
b) Solve for t:
s= vt + 1/2at^2
315= 48.3t + 1/2(3.7)t^2
315= 48.3t + 1.85t^2
1.85t^2 + 48.3t - 315 = 0
t = 5.40s
v^2 = 2 x 3.7 x 315
v = 48.3 m/s
b) Solve for t:
s= vt + 1/2at^2
315= 48.3t + 1/2(3.7)t^2
315= 48.3t + 1.85t^2
1.85t^2 + 48.3t - 315 = 0
t = 5.40s
2016-06-08 5:29 am
a. d = (1/2)at^2
315 = (1/2)(3.7)t^2 <<< solve for t
t = sqrt [2(315)/3.7] = 13.05 sec
---------------------
v = at = 3.7(13.05) = 48.285 m/s
b. see above ... 13.05 sec.
315 = (1/2)(3.7)t^2 <<< solve for t
t = sqrt [2(315)/3.7] = 13.05 sec
---------------------
v = at = 3.7(13.05) = 48.285 m/s
b. see above ... 13.05 sec.
2016-06-08 5:28 am
(a) 2gy = v^2 =>
v = sqrt(2*3.7 m/s^2*315 m) = 48.6 m/s.
(b) The average speed is 24.3 m/s, so the time is (315 m)/(24.3 m/s) = about 13.0 seconds.
v = sqrt(2*3.7 m/s^2*315 m) = 48.6 m/s.
(b) The average speed is 24.3 m/s, so the time is (315 m)/(24.3 m/s) = about 13.0 seconds.
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