A rock is thrown directly upward from the edge of the roof of a building that is 37.2 meters tall. The rock misses the building on its way down, and is observed to strike the ground 4.00 seconds after being thrown. Neglect any effects of air resistance. With what speed was the rock thrown?
since h(t) = h0 + v0*t - 4.9t^2, just solve for v in
37.2 + 4v - 4.9*16 = 0
To find the speed with which the rock was thrown, we will use the equation of motion that relates the height, time, and initial velocity of a projectile.
The equation we will use is:
h = ut + (1/2)gt^2
Where:
h = height (37.2 meters)
u = initial velocity (unknown)
t = time (4.00 seconds)
g = acceleration due to gravity (-9.8 m/s^2)
Since the rock is thrown directly upward, the initial velocity will be in the opposite direction of gravitational acceleration. Therefore, u = -u0, where u0 is the initial velocity.
First, let's calculate the time it takes for the rock to reach its highest point. At the maximum height, the velocity becomes zero.
Using the kinematic equation:
v = u + gt
Substituting v = 0, and g = -9.8 m/s^2, we get:
0 = u - 9.8t1
t1 = u / 9.8
Now, let's calculate the time it takes for the rock to hit the ground. The total time of flight is the sum of the time to reach the maximum height and the time to fall back down:
t2 = t - t1
Now, we can substitute this value of t1 into the equation for height and solve for the initial velocity u:
h = -u0t2 + (1/2)g(t2)^2
Substituting the known values:
37.2 = -u0(t - u / 9.8) + (1/2)(-9.8)(t - u / 9.8)^2
Simplifying and rearranging the equation, we can solve for u0:
37.2 = -u0t + u0^2 / 9.8 + (-4.9)t^2 + ut
Rearranging the equation to isolate u0^2, we get:
u0^2 / 9.8 = 4.9t^2 + ut - 37.2
Now, we can substitute the known values of t and solve for u0:
u0^2 = 9.8(4.9t^2 + ut - 37.2)
Take the square root of both sides:
u0 = √(9.8(4.9t^2 + ut - 37.2))
Finally, substitute the value of t and calculate the initial velocity u0.
To find the initial velocity or speed at which the rock was thrown, we can use the equation of motion:
v = u + gt
where:
v = final velocity (0 m/s, as the rock hits the ground)
u = initial velocity (unknown)
g = acceleration due to gravity (-9.8 m/s², assuming downward direction)
t = time (4.00 seconds)
Since the rock is thrown directly upward, the initial velocity is in the opposite direction of gravity, so we can take it as positive.
Plugging the given values into the equation, we have:
0 = u + (-9.8 m/s²)(4.00 s)
0 = u - 39.2 m/s²
Rearranging the equation to solve for u:
u = 39.2 m/s²
Therefore, the rock was thrown with an initial speed of 39.2 m/s.