# Projectile Motion Problem - Solving for Maximum Height and Landing Distance

• Melawrghk
In summary, the projectile motion problem has been solved by finding the time it will take for the skier to hit the ground from her highest point, and then plugging that into the equation for vertical distance travelled.
Melawrghk
Projectile motion problem - SOLVED

## Homework Statement

This is one of the last parts of a bigger problem, but I get the first parts. So:

The projectile motion describes a skier's jump. The initial velocity is 21.469m/s at 20 degrees to the horizontal. Find the maximum height the skier reaches (relative to the take off point), find distance 'd' down the slope where the skier lands.

## The Attempt at a Solution

First I wrote the equations:
Vx = 21.469cos(20)
Vy= 21.469sin(20)-gt

Then, I found the time it took for the skier to reach the top:
Vy=21.469sin(20)-9.81t
9.81t=21.469sin(20)
t= 0.7485 seconds

Then I found the 'h':
ds/dt = 21.469sin(20)-9.81t
*integrating...*
s=21.469sin(20)t-4.905t2
s(0.7485) = 2.7481 m

Then I drew this:

On it, I calculated the 30.2 value by plugging in 0.7485*2 into an x-direction distance equation:
Sx = 0.7485*2 x 21.469cos20 = 30.2
I then found the 17.44 using just trig.

So now, I started to treat the big triangle (described by p and q) as a similar triangle with sides 30.2&17.44 and developed a relationship:
q/p = 17.44/30.2
q=0.577p

Next, I expressed d in terms of p and q:
d2=p2+q2
d2= 1.3329p2
and thus, p=0.866d
I performed similar operations and got:
q=0.5d

Then time for the skier to go the whole horizontal distance:
0.886d = 21.469cos(20)*t
t=0.04293d
And thus the time it will take for the skier to hit the ground from her highest point is:
t=0.0429d-0.7485

Now, we can plug in that equation into the equation of vertical distance travelled:
h+0.5d = 21.469sin(20)(0.0429d-0.7485)-4.905(0.04293d-0.7485)2
I can simplify that and try to find d (from discriminant and things) but I end up with a negative discriminant... Help?

Last edited:
You might be better served to solve it more simply.

y = Vy*t - 1/2*g*t2

And because of the slope y = -x*tan30

and x = Vx*t

Where y intersects with the slope is where he lands.

-Vx*t*tan30 = Vy*t - 1/2*g*t2

Solve for t and plug it into either equation.

Wow, thanks! That's so much easier and makes a lot more sense! :)

## 1. What is projectile motion?

Projectile motion refers to the motion of an object that is projected into the air and then moves under the influence of gravity alone. Examples of projectile motion include throwing a ball, shooting a projectile from a cannon, and jumping off a diving board.

## 2. What are the key components of a projectile motion problem?

The key components of a projectile motion problem include the initial velocity of the object, the angle at which it is projected, the acceleration due to gravity, and the time the object is in the air.

## 3. How do you solve for the maximum height and range in a projectile motion problem?

To solve for the maximum height in a projectile motion problem, you can use the equation h = (v₀sinθ)²/2g, where v₀ is the initial velocity and θ is the angle of projection. To find the range, you can use the equation R = (v₀² sin2θ)/g.

## 4. Can a projectile's initial velocity and angle of projection affect its trajectory?

Yes, the initial velocity and angle of projection can greatly affect a projectile's trajectory. A higher initial velocity will result in a longer range, while a smaller angle of projection will cause the projectile to travel higher but with a shorter range.

## 5. How does air resistance affect projectile motion?

Air resistance can affect projectile motion by slowing down the object's motion and causing it to deviate from its intended trajectory. This is more significant for objects with larger surface areas, such as a feather or a parachute, as they experience more air resistance compared to smaller objects like a bullet.

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