a boy uses a sling shot to launch a rock upward with a velocity of 30m/s how long does it take to reach a height of 20 m what maximum height does the rock reach, what is the velocity of the rock at this height, what is the velocity of the rock just before it lands
h = ½ g t² + 30 t = -4.9 t² + 30 t
substitute 20 for h to find the time
... use the quadratic formula
the time of max h is on the axis of symmetry of the parabola
... t = -b / 2a = -30 / -9.8
... plug in the time to find h max
the velocity is zero at max h
the impact velocity equals the launch velocity (but opposite direction)
... this ignores the height of the launcher and air resistance
To solve this problem, we can use the equations of motion. Let's first find the time it takes for the rock to reach a height of 20 m.
1. Begin by remembering that the initial velocity (u) of the rock is 30 m/s and the acceleration (a) due to gravity is approximately 9.8 m/s^2 (assuming no other external factors).
2. Use the following kinematic equation to find the time (t) it takes for the rock to reach a height of 20 m:
h = u*t + (1/2)*a*t^2
Substituting the given values:
20 = 30*t - (1/2)*9.8*t^2
3. Rearrange the equation to solve for t in terms of a quadratic equation:
-4.9t^2 + 30t - 20 = 0
4. Solve this quadratic equation using a method of your choice (factoring, completing the square, or quadratic formula) to find the values of t. In this case, it can be factored as:
-4.9(t^2 - 6.12t + 4) = 0
Factoring further gives:
-4.9(t - 0.82)(t - 5.18) = 0
So, t = 0.82 s (ignore the negative value) or t = 5.18 s.
Therefore, it takes approximately 0.82 seconds for the rock to reach a height of 20 m.
Next, let's find the maximum height the rock reaches and its velocity at this height.
1. We know that at the maximum height, the vertical velocity (v) becomes zero, and the only force acting on the rock is the force of gravity pulling it downward.
2. Use the following equation to find the maximum height (H) the rock reaches:
v^2 = u^2 - 2aH
Here, v = 0, u = 30 m/s, and a = 9.8 m/s^2.
Solving for H:
0 = 30^2 - 2*9.8*H
H = (30^2) / (2*9.8)
Therefore, the maximum height the rock reaches is approximately 46.94 meters.
3. To find the velocity at this height, we can use the equation:
v = u - a*t
Substituting the given values:
v = 30 - 9.8*t
At the maximum height, t = 0.82 s
v = 30 - 9.8*0.82
Therefore, the velocity of the rock at the maximum height is approximately 21.436 m/s.
Finally, let's find the velocity of the rock just before it lands.
1. Recall that just before the rock lands, the height (h) is zero, and the final velocity (v) is what we need to find.
2. Using the equation of motion:
h = u*t - (1/2)*a*t^2
Substituting the given values:
0 = 30*t - (1/2)*9.8*t^2
Solve this equation to find the time it takes for the rock to land. In this case, the solutions are t = 0 and t = 6.12 seconds. Since the rock started from the ground, t = 0 can be ignored.
3. To find the velocity of the rock just before it lands, use the equation:
v = u - a*t
Substituting the given values:
v = 30 - 9.8*6.12
Therefore, the velocity of the rock just before it lands is approximately -35.34 m/s (negative sign indicates downward direction).
To summarize:
- The time it takes for the rock to reach a height of 20 m is approximately 0.82 seconds.
- The maximum height the rock reaches is approximately 46.94 meters.
- The velocity of the rock at this maximum height is approximately 21.436 m/s.
- The velocity of the rock just before it lands is approximately -35.34 m/s.