# Projectile Motion Jump Question

• yoyo16
In summary, the person jumps 8.9 metres and their takeoff speed was 15 metres per second. The angle relative to the ground at takeoff was 0.5 degrees.
yoyo16

## Homework Statement

A person jumps 8.9m. The person's takeoff speed was 15m/s. What is the angle relative to the ground at takeoff?

## The Attempt at a Solution

I found t using the kinematic formula to be 0.5s using the formula. d=vit+1/2at^2. Can someone tell explain to me what step to do next. Thanks

I assume the jump distance stated is horizontal only.

You can't use that kinematic formula straight out the way you did. You have to break it into two different directions. So you will have two formulas but three unknowns: t, dy, and θ. You get the 3rd equation using a trigonometric or geometric equation.

ummm, i attempted it so basically i used y=1/2at^2. so imagine you're at the peak of 8.9 meters, use that equation to calculate the amount of air time before reaching the air. after that use that t to plug in the vertical kinematic equation:
y= y(knot) + v(knot)t - 1/2at^2. plug 8.9 in. For v(knot) set it equals to v(knot)sin(theta). Then just find theta.
it's going to be parabolic path, therefore the time it takes to get into the air is the same as the time it takes to fall.

I'm assuming that he's jumping 8.9 vertically?so the math I just articulated is for vertical distance

Last edited:
Ummm...well you have to clarify is the person jumping 8.9m high (y direction) or 8.9m horizontally (x direction)?

I'm sorry about that. 8.9m was for the horizontal distance.

paisiello2 said:
Ummm...well you have to clarify is the person jumping 8.9m high (y direction) or 8.9m horizontally (x direction)?

If someone could jump 8.9 m vertically, he would be the president for life of Phi Slamma Jamma.

yoyo16 said:
I'm sorry about that. 8.9m was for the horizontal distance.
Right, so then you can't use the equation the way you did.

Can you set up the 3 different equations?

ok it's pretty simple, you set up 2 equations of T(air time)
so the first one is t=(vsin(theta))/g and they other one is t=x/(vcos(theta))
set them equal to each other, plug v and x in and find theta, that's it

Edit: sry, for the first one just times it by 2 so t=2(whatever is there)

Last edited:
You'll need one more trigonometric identity equation to find theta.

Equation for trajectory says $$x = \frac {{v_{0}}^2\sin 2\theta}{g}$$
Plug in your unknowns from there.

Thanks guys for the help!

## 1. What is projectile motion?

Projectile motion is the motion of an object (called a projectile) through the air, typically with a curved path, under the influence of gravity.

## 2. How is projectile motion different from regular motion?

In regular motion, an object moves in a straight line with a constant velocity. In projectile motion, the object follows a curved path due to the influence of gravity.

## 3. What is the equation for calculating the horizontal distance of a projectile?

The equation for calculating the horizontal distance of a projectile is d = v0 * t * cosθ, where d is the horizontal distance, v0 is the initial velocity, t is the time, and θ is the angle of launch.

## 4. How does the angle of launch affect the projectile's trajectory?

The angle of launch affects the projectile's trajectory by determining the shape and range of the path. A higher angle will result in a longer, more parabolic path, while a lower angle will result in a shorter, more linear path.

## 5. Can you predict the landing point of a projectile?

Yes, the landing point of a projectile can be predicted by using the equations of projectile motion and knowing the initial conditions (such as angle of launch, initial velocity, and height). However, external factors such as air resistance and wind can also affect the landing point.

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