# Ball launched off a cliff at negative angle

• picklepie159
In summary, the ball is thrown off a cliff at a negative angle and it lands 120 meters away. The ball was thrown at -15 degrees horizontal.
picklepie159
Ball launched off a cliff at negative angle...

## Homework Statement

So a ball is launched off a cliff 60 meters high
It is thrown at -15 degrees horizontal, and lands 120 meters away.

Find the V-initial, V-final, Time in air, and Angle at which it hits the ground

## Homework Equations

inverse tan(75)= Vx/Vy
D= VT
h=1/2g t^2
Vx= cosin(75) x V-initial
Vy=sin(75) x V-initial

## The Attempt at a Solution

I know that there's a policy that I have to try to work on the problem first before i get help, but man, I am lost. This is what I can make out.

I switched everything to positive so now its falling off the cliff at 75 degrees.
120 meters = Vx T

That's all. A little ray of guidance would be a Godsend.

PS
Would the Velocity-y be V-initial +34.6 m/s?
Because when something falls 60 meters from rest, the final velocity is 34.6 meters.

Last edited:

What do you mean you switched everything to positive?

switching everything positive is something i don't understand in your solution...

so my approach would be ...
Range = v*cos(-15) * t

Height = v*sin(-15)t + 1/2*g*t^2

now ... you get 2 equations in v and t and there are 2 variables... so you can solve for both...do you get it?

By switching to positive, I ust meant I flipped it upside down- the movement towards the ground is now the positive direction. Then, the angle it was thrown at is 75 degrees instead of -15. (Does that work out?)

It just makes stuff easier- no fiddly negative directions to deal with.

then don't you think the time of flight changes in both cases?

Really sorry for digging up a super old thread but I'm trying to solve a similar problem right now.

As The Legend suggested, I've solved for v and t using those equations. First, I define v in terms of R and t using the R = v * cosθ * t

v = R/(cos(θ) * t)

Then, I substitute that into the height equation...

H = (R/(cos(θ) * t)) * sin(θ) * t + 1/2 * g * t^2

and solve for t...

t = √((R * tan(θ) + H)/(g * 1/2))

Solve it using my values, then substitute that back into the velocity derivation above...

This works for angles [0..90] but anything below the horizon just craps out due to a -ve √

Where am I going wrong? This is driving me insane!

Just wanted to clarify something, the reason I add H is because I actually got

--H in my particular case..

Nightro said:
t = √((R * tan(θ) + H)/(g * 1/2))

Actually... I'm wrong.. it works for the inputs given at the top of this thread :S

I think what might actually be happening in my case is my launch angle is too much!

That is to say... the angle that is a direct line to the target location is more than the angle I'm trying to launch it at (e.g.)

Angle from Launch to Target: -10
Launch Angle: -20

That would mean its impossible, right? And why I'm getting these imaginary numbers :)

## What is the initial velocity of the ball launched off a cliff at a negative angle?

The initial velocity of the ball is the speed and direction at which the ball is launched. It can be calculated using the equation V0 = V * cos(theta), where V is the initial speed and theta is the launch angle.

## How does the launch angle affect the trajectory of the ball?

The launch angle determines the direction in which the ball will travel. A negative launch angle means the ball is launched downwards, and its trajectory will follow a curved path. The steeper the angle, the more vertical the trajectory will be.

## What factors affect the distance the ball travels after being launched off a cliff at a negative angle?

The factors that affect the distance the ball travels include the initial speed of the ball, the launch angle, and the force of gravity. The higher the initial speed and the shallower the launch angle, the farther the ball will travel. However, the force of gravity will decrease the distance traveled.

## How long will it take for the ball to reach the ground?

The time it takes for the ball to reach the ground can be calculated using the equation t = (2 * V0 * sin(theta)) / g, where V0 is the initial speed, theta is the launch angle, and g is the acceleration due to gravity.

## What happens to the ball after it reaches the ground?

Once the ball reaches the ground, it will continue to roll or bounce depending on the surface it lands on. The force of gravity will also act on the ball, causing it to eventually come to a stop.

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