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 = ?
3 answers
t = 0.5s
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 :-)
"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.
2 answers