Projectile Motion stone dropped

In summary, the conversation discusses using physics to solve two problems, one involving a falling stone and the other involving a jumping kangaroo. The necessary variables and formulas are provided, and the solution requires substituting the values into the formula to find the desired answer.
  • #1
bkoiskewl1
7
0
1)If a stone dropped (not thrown) from a bridge takes 4.2 s to hit the water, how high is the rock-dropper? Ignore friction.

2)A kangaroo can jump straight up about 2.0 m. What is its takeoff speed?


I suck at physics so yeah...
 
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  • #2
bkoiskewl1 said:
1)If a stone dropped (not thrown) from a bridge takes 4.2 s to hit the water, how high is the rock-dropper? Ignore friction.

2)A kangaroo can jump straight up about 2.0 m. What is its takeoff speed?


I suck at physics so yeah...

You don't suck at physics, you just need to get in the right frame of mind:biggrin:

Check out this thread: https://www.physicsforums.com/showthread.php?t=110015" for useful formulae.

List all the variables that you know and the variables that you need to know., then find a formula that includes everything you know and one thing you don't. Substitute into the formula and you will get the answer you seek.

For your first question, you know several things:
V0 (initial velocity) = 0
a (acceleration) = g (gravity, or 9.8 m/s2)
t (time) = 4.2s
x0 (starting position) = 0

what you are looking for is
x (finishing distance)

Find the formula, plug the numbers in, find your answer.

Your second question takes a little bit more thinking, but try it out and let me know how far you get.
 
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  • #3


1) To determine the height of the rock-dropper, we can use the equation h = 1/2gt^2, where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time it takes for the stone to hit the water. Plugging in the given time of 4.2 seconds, we get h = 1/2(9.8)(4.2)^2 = 86.94 meters. This means that the rock-dropper was standing 86.94 meters above the water when they dropped the stone.

2) To calculate the takeoff speed of a kangaroo, we can use the equation v = √(2gh), where v is the takeoff speed, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the jump. Plugging in the given height of 2.0 meters, we get v = √(2(9.8)(2)) = 6.26 m/s. This means that the kangaroo needs to have a takeoff speed of 6.26 m/s in order to jump 2.0 meters straight up.
 

1. What is projectile motion and how does it apply to a stone being dropped?

Projectile motion is the motion of an object through the air that is subject to only the force of gravity. When a stone is dropped, it is considered a projectile and will follow a parabolic path due to the force of gravity acting on it.

2. What factors affect the trajectory of a stone dropped in projectile motion?

The factors that affect the trajectory of a stone dropped in projectile motion include the initial velocity, angle of release, and air resistance. These factors determine the shape and distance of the parabolic path the stone will follow.

3. How does the height from which a stone is dropped impact its motion?

The height from which a stone is dropped will impact its motion by changing the initial velocity and angle of release. The higher the drop height, the greater the initial velocity and the steeper the angle of release, resulting in a longer and narrower parabolic path.

4. What is the equation for calculating the maximum height reached by a stone in projectile motion?

The equation for calculating the maximum height reached by a stone in projectile motion is h = (v02sin2θ)/2g, where h is the maximum height, v0 is the initial velocity, θ is the angle of release, and g is the acceleration due to gravity (9.8 m/s2).

5. Can the range of a stone dropped in projectile motion be increased by changing the angle of release?

Yes, the range of a stone dropped in projectile motion can be increased by changing the angle of release. The optimal angle for maximum range is 45 degrees. Any angle less than 45 degrees will result in a shorter range, and any angle greater than 45 degrees will result in a shorter range due to the steeper angle of descent.

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