How Does Projectile Motion Work When a Ball is Thrown from a Cliff?

[wolves5]

A ball is thrown from the brink of a cliff 10 m high with an initial velocity of v0 = 4 m/s at an angle q0 = 31° above the horizontal.

a) How long does it take to land?
I did 4sin(31)-.5(-9.8)t^2, but that is not the right answer.

b) How far from the base of the cliff does it land?
I might be able to solve this one if I knew the time from part a. Can I just use d=vit + 0.5at^2?
c) What is its maximum height above the base of the cliff?
I don't know what to do for this one.

 

[Yuqing]

For a) the ball has to travel 10 meters in the opposite direction of the initial velocity. If we define up as positive, then we get

-10 = 4sin(31)t - 0.5gt²

using the quadratic formula to solve for t will give you the time.

Try b) once you have the time for a).

For c) there are several ways to approach this. The velocity at the top of the trajectory has a vertical velocity of 0. Using this fact, you should be able to solve for the height.

 

[wolves5]

I got a and b. For c, I used the equation: y=(vosin(theta))t -0.5gt^2. I did (4sin(31)(1.654) - 0.5(-9.81)(1.654^2). But its not the right answer. I don't know which equation to use know.

 

[ideasrule]

I got a and b. For c, I used the equation: y=(vosin(theta))t -0.5gt^2. I did (4sin(31)(1.654) - 0.5(-9.81)(1.654^2). But its not the right answer. I don't know which equation to use know.

That would be right if you used the right value for time. So how did you get 1.654 s for the time?

 

[wolves5]

I used the time from part a. Isn't that right?

 

[Yuqing]

The time for part a is the time it takes to reach the ground. You need the time it takes for the ball to reach zero vertical velocity.

 

[wolves5]

Would that be 2.5 seconds? I did Distance / Velocity. (10/4=2.5 sec). When I plugged that in, I didnt get the right answer.

 

[Yuqing]

Well, you have an initial vertical velocity of 4sin(31). You have an acceleration of -9.81m/s². You can simply use a = (v₂-v₁)/t to solve for time.

 

[wolves5]

Ok I did that and got 0.4082 seconds as my time. Then I did 4sin(31)(0.4082) - 0.5(-9.81)(0.4082^2). I got 0.02365m, but that's not right.

 

[Yuqing]

How did you get 0.4082 seconds?

 

[wolves5]

I used the equation you gave me. a = (v2-v1)/t

So -9.8=(0-4)/t. I got t=0.4082 seconds

 

[Yuqing]

Notice that throughout the discussion, I always use vertical velocity. Gravity only acts in the vertical direction. You need to use 4sin(31) instead of 4.

 

[wolves5]

Ok. So I did that and I got 0.210 seconds. Then I plugged it into the equation: 4sin(31)(0.210)-0.5(-9.81)(0.210^2). I got 0.649m, but that's not right.

 

[Yuqing]

The equation for distance of a uniformly accelerated object is

d = v₀t + 0.5at²

we normally write the second term as -0.5gt² because we take gravity as negative. You essentially did that twice for your equation. Switch the sign of the second term and it should give you the proper answer. (Don't forget to add the distance to the bottom of the cliff.)