# Solving Projectile Motion Problem - Tips Needed

• Jennifer K
In summary, the problem involves a projectile of mass 33.6 grams being thrown from a height of 1 meter at an angle of 60 degrees, landing 10 meters away. The peak vertical distance and initial velocity (Vi) are asked for. The relevant equations are provided, and the solution involves finding the time it took to travel 10 meters and reach the maximum height, as well as using the equation v^2=u^2-2gh_{max} to find the maximum height.
Jennifer K

## Homework Statement

Hello i'd apprcieate it if someone could help me with this projectile motion problem.

A projectile off mass 33.6 grams is thrown from a hight of 1 meter of the ground at an angle of 60 degress, it lands 10 meters away from the point it was thrown.

What is the peak vertical distance?
What is Vi?

## Homework Equations

(1) V = U+at^2

(2) s = 1/2 (U+V)t

(3) s = Ut + 1/2 at^2

(4) V^2 = U^2 +2as

## The Attempt at a Solution

I tried to do it in two parts:

(1)

U = ViSIN60

V = 0

a = -9.81 m/s^2

s1 = ?

t = ?

(2)

U = 0

V = ?

a = -9.81

Stotal = S1 + 1 meter

t2 = ?

and then for the horizontal:

U = ViCOS60

V = ?

a = 0

S = 10 meters

ttotal = t1 + t2

i'm really stuck if anyone can give me any tips i will really appreciate it. thank youuuuuu :) :) :)

Jenny

The time it took to travel 10 meters is t=10/vcos60. This is the same time it took to reach the max height, h, and fall h+1 meters.
Individually find the time it took the projectile to reach the max height, and then fall h+1 meters. This time is equal to t.
You can use $$v^2=u^2-2gh_{max}$$ to get the max height.

This should give you enough equations to solve for the variables.

, it looks like you have made a good start on this problem. To find the peak vertical distance, you will need to use equation (3) with U = ViSIN60, V = 0, and a = -9.81 m/s^2. You can then solve for s, which will give you the peak vertical distance.

As for finding Vi, you will need to use equation (4) with V = 0, a = -9.81 m/s^2, and s = 1 meter. This will give you the initial velocity in the vertical direction.

For the horizontal motion, you can use the same approach as you did for the vertical motion. Use equation (3) with U = ViCOS60, V = ?, and a = 0 to solve for the time it takes for the projectile to travel 10 meters horizontally.

Remember to use the same initial velocity for both the horizontal and vertical motions, since the projectile is thrown at an angle. Let me know if you need any further assistance. Good luck!

## 1. What is projectile motion?

Projectile motion is the motion of an object that is launched into the air and is subject to only the force of gravity. This type of motion can be seen in everyday phenomena such as throwing a ball or shooting a projectile.

## 2. What are the factors that affect projectile motion?

The factors that affect projectile motion are the initial velocity, angle of launch, and the force of gravity. These factors determine the trajectory and distance traveled by the projectile.

## 3. Can you calculate the maximum height and range of a projectile?

Yes, the maximum height and range of a projectile can be calculated using the equations for projectile motion, which take into account the initial velocity, angle of launch, and gravitational force.

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

Air resistance can affect projectile motion by slowing down the projectile and changing its trajectory. This effect is more prominent for objects with larger surface areas and lower velocities.

## 5. What are some real-life applications of projectile motion?

Projectile motion is used in a variety of fields, including sports, engineering, and physics. Examples include throwing a ball in sports, launching rockets into space, and calculating the trajectory of a bullet.

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