Projectile Motion: Boulder from Cliff to Plain

[jackleyt]

1. A 76.0 kg boulder is rolling horizontally at the top of a vertical cliff that is 20.0 m above the surface of a lake, as shown in figure below. (Intro 1 figure) The top of the vertical face of a dam is located 100 m from the foot of the cliff, with the top of the dam level with the surface of the water in the lake. A level plain is 25.0 m below the top of the dam.

1)What must the minimum speed of the rock be just as it leaves the cliff so that it will travel to the plain without striking the dam?

2)How far from the foot of the dam does the rock hit the plain?

Homework Equations

Equations of Motion, Projectile Motion

The Attempt at a Solution

I don't know how to incorporate the mass of the rock in the equation. I don't know where to start!

 

[hage567]

I don't know how to incorporate the mass of the rock in the equation. I don't know where to start!

Perhaps you don't need the mass? How would you approach the problem if you didn't know what the mass was?

Sometimes you don't always use all the information given to you in the question.

 

[baba89]

I would first start by defining the problem and identifying the relevant variables. In this case, we are dealing with projectile motion, which is the motion of an object that is launched into the air and moves under the influence of gravity. The key variables in this problem are the initial velocity, the angle of launch, the height of the cliff, and the distance to the dam and plain.

To solve for the minimum speed of the rock, we can use the equation for the horizontal distance traveled by a projectile, which is given by:
x = v*cosθ*t

Where:
x = horizontal distance
v = initial velocity
θ = angle of launch
t = time

Since the rock is rolling horizontally, we can assume that the angle of launch is 0 degrees. We also know that the distance to the plain is 100m from the foot of the cliff, and the height of the cliff is 20m. Therefore, the equation becomes:
100 = v*cos0*t
100 = v*t

Next, we need to find the time it takes for the rock to reach the plain. This can be done by using the equation for the vertical distance traveled by a projectile, which is given by:
y = v*sinθ*t - (1/2)*g*t^2

Where:
y = vertical distance
v = initial velocity
θ = angle of launch
t = time
g = acceleration due to gravity (9.8 m/s^2)

Since the rock starts at a height of 20m and we want it to reach a height of 25m (the plain), we can set the equation equal to 25 and solve for t:
25 = v*sin0*t - (1/2)*9.8*t^2
25 = 0*t - 4.9*t^2
t = √(25/4.9) = 2.55s

Now, we can substitute this value for t in the first equation to solve for v:
100 = v*2.55
v = 100/2.55 = 39.22 m/s

Therefore, the minimum speed of the rock just as it leaves the cliff is 39.22 m/s.

To find the distance from the foot of the dam to where the rock hits the plain, we can use the same equation for horizontal distance:
x = v*cosθ*t
Since the angle of launch is