A rock is thrown vertically upward with a speed of 11.0 m/s from the roof of a building that is 70.0 m above the ground. Assume free fall.
In how many seconds after being thrown does the rock strike the ground?
GOT THIS!!!! t=5.07s
What is the speed of the rock just before it strikes the ground?
I HAVE NO IDEA HOW TO DO THIS!!!
Let's check this
s(t)=(-1/2)gt^2+v_0t+s_0
s_0=70m, v_0=11m/s, g=9.8m/s
We want t for s(t)=0
0=(-1/2)9.8t^2+11*t+70
Solving for t 0=-4.9t^2+11t+70
t= -11/9.8 +/-sqrt(121 - 4*(-4.9)70)/-9.8
t = 5.07 so the time is correct
v(t)=-gt+v_0 at t=5.07 v=-9.8*5.07+11=-38.7m/s downward
Well, it looks like before the rock strikes the ground, it's going to have a serious need for speed. The speed of the rock just before it strikes the ground is approximately 38.7 m/s, but in a downward direction. So, it's safe to say that it won't be breaking any land speed records. But hey, at least it's got a one-way ticket to gravity town!
The speed of the rock just before it strikes the ground is approximately 38.7 m/s downward.
To find the speed of the rock just before it strikes the ground, we can use the equation for velocity in free fall. The equation is v(t) = -gt + v0, where v(t) is the velocity at time t, g is the acceleration due to gravity, and v0 is the initial velocity.
In this case, g is -9.8 m/s (negative because it's directed downwards), v0 is 11.0 m/s (upwards), and the time we found is t = 5.07 s.
Plugging in the values, we get v(5.07) = -9.8 * 5.07 + 11 = -38.7 m/s (downward). So, the speed of the rock just before it strikes the ground is 38.7 m/s downward.