A rock is thrown vertically with a velocity of 21 m/s from the edge of a bridge 41 m above the river. How long does the rock stay in the air?
V = 21 m/s
d = 41 m
a = -9.80 m/s^2
t = ?
Start by translating the question into "physics language:"
"If the rock is thrown up at time t=0, what is the time, t1, when the rock hits the water.
You can use the usual equation
x = 1/2 a t^2 + v0 t + x0.
v0 = 21 m/sec and x0=41 meters (taking x=0at the water surface)
That works for all the time that the rock is in the air. You want the particular case when x=0, so plug in and solve that quadratic :-)
On my homework a rock is thrown vertically upward with a velocity of 21 m/s from the edge of a bridge 42 m above a river. How long does the rock stay in the air? My question is only 1 meter off your distance but the answer is 5.8 seconds. No idea how they got 5.8 seconds because I got 4 when i did my distance formula, so 0.5 can't be right.
To find the time the rock stays in the air, we can use the equations of motion for vertical motion.
The equation that relates the final velocity (V), initial velocity (U), acceleration (a), and time (t) is:
V = U + at
Since the rock is thrown vertically upwards, the final velocity at the highest point is 0 m/s, and the acceleration is -9.8 m/s^2 due to gravity. The initial velocity is 21 m/s in the upward direction. We can plug these values into the equation and solve for time (t).
0 = 21 + (-9.8)t
To isolate t, we can rearrange the equation:
9.8t = 21
t = 21 / 9.8
Calculating the value, we find that t is approximately 2.14 seconds.
Therefore, the rock stays in the air for about 2.14 seconds.