a person uses a slingshot to launch a rock upward with a velocity of 30m/s how long does it take to reach a height of 20m 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
upward motion:
h'=vi*t=1/2 * 9.8 t^2
h'=20, vi=30, solve for t using quadratic formula
now with t, v=vi-g*t, put vi=20, and t from above, and you have v.
Max height:
initial KE=final PE
1/2 m vi^2=mgh solve for h, the max height.
Velocity just before landing = negative of launch velocity.
To answer these questions, we need to use the basic equations of motion. Let's break down each question one by one.
1. How long does it take to reach a height of 20m?
We can use the equation of motion:
h = u*t + (1/2)*a*t^2,
where h is the height, u is the initial velocity, a is the acceleration (which is the acceleration due to gravity, approximately 9.8 m/s^2), and t is the time.
Rearranging the equation, we get:
t = (sqrt(2*h/a)),
Substituting the given values:
t = (sqrt(2*20/9.8)) ≈ 2.02 seconds.
Therefore, it takes approximately 2.02 seconds for the rock to reach a height of 20m.
2. What is the maximum height the rock reaches?
The maximum height occurs when the vertical velocity becomes zero. At that point, the only force acting on the rock is gravity. We can use the equation:
v = u + a*t,
where v is the final velocity. At the maximum height, the final velocity becomes zero. Rearranging the equation:
t = -u / a.
Substituting the given values:
t = -30 / -9.8 ≈ 3.06 seconds.
To find the maximum height, we can use the equation of motion with this time:
h = u*t + (1/2)*a*t^2,
h = 30*3.06 + (1/2)*(-9.8)*(3.06)^2,
h ≈ 45.84 meters.
Therefore, the maximum height the rock reaches is approximately 45.84m.
3. What is the velocity of the rock at this height?
To find the velocity at the maximum height, we can use the equation we obtained previously:
v = u + a*t.
Substituting the given values:
v = 30 - 9.8*3.06 ≈ -0.38 m/s.
Therefore, the velocity of the rock at the maximum height is approximately -0.38 m/s. The negative sign indicates that the velocity is in the opposite direction of the initial velocity.
4. What is the velocity of the rock just before it lands?
We know the time it takes to reach the maximum height, which is 3.06 seconds. Doubling this time will give us the total time the rock is in the air. Therefore, just before landing, the time would be approximately 2 * 3.06 = 6.12 seconds.
Using the equation of motion:
v = u + a*t,
Substituting the time and initial velocity, we get:
v = 30 - 9.8*6.12 ≈ -34.78 m/s.
Therefore, the velocity of the rock just before it lands is approximately -34.78 m/s. Again, the negative sign indicates that the velocity is in the opposite direction of the initial velocity.