A stone is launched straight up by a slingshot. Its initial speed is 20.6 m/s and the stone is 1.59 m above the ground when launched. How high above the ground does the stone rise? How much time elapses before the stone hits the ground?
hf=hi+vi*t-4.9t^2
hf=0, hi=1.59, vi given, solve for time t in the air.
how high? at the top, vf=0
vf^2=vi^2-2g*height
solve for height.
To find the maximum height the stone reaches, you can use the kinematic equation relating displacement, initial velocity, final velocity, acceleration, and time:
\[v_f^2 = v_i^2 + 2a \Delta y\]
Initially, the stone is launched straight up, so its final velocity at the highest point is zero. The acceleration, in this case, is due to gravity and will be -9.8 m/s² (negative because gravity acts in the opposite direction of the stone's motion). The initial velocity is 20.6 m/s, and the displacement is the difference in height between the highest point and the initial position, which is 1.59 m.
Plugging the values into the equation:
\[0 = (20.6)^2 + 2(-9.8) \Delta y\]
Simplifying:
\[0 = 424.36 - 19.6 \Delta y\]
Rearranging the equation:
\[\Delta y = \frac{424.36}{19.6}\]
Solving for \(\Delta y\):
\[\Delta y = 21.68 \, \text{m}\]
So the stone rises approximately 21.68 meters above the ground.
To find the time it takes for the stone to hit the ground, we can use the equation:
\[\Delta y = v_i t + \frac{1}{2} a t^2\]
Since we know that at the highest point the displacement is 21.68 meters and the acceleration is -9.8 m/s², we can plug in the values:
\[21.68 = 20.6 t + \frac{1}{2} (-9.8) t^2\]
Simplifying:
\[21.68 = 20.6 t - 4.9 t^2\]
Rearranging the equation:
\[4.9 t^2 - 20.6 t + 21.68 = 0\]
This is a quadratic equation that can be solved using the quadratic formula. The standard quadratic formula is:
\[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
In this case, a = 4.9, b = -20.6, and c = 21.68. Plugging in the values:
\[t = \frac{-(-20.6) \pm \sqrt{(-20.6)^2 - (4)(4.9)(21.68)}}{2(4.9)}\]
Now, solve for t.