# Projectile Problem No air resistance(At what velocity was object thrown?)

• 1irishman
In summary, the conversation discusses solving for the initial velocity of an object thrown into the air at an angle of 35 degrees to the horizontal. The vertical and horizontal components of the motion are analyzed using equations for displacement, acceleration, and time. The final answer is found to be 79.1 m/s by solving for the initial vertical velocity and using trigonometry to find the total velocity. Other methods, such as using the maximum height reached by the object, could also be used to solve for the initial velocity.
1irishman

## Homework Statement

An object is thrown into the air from the ground at an angle of 35deg to the horizontal. If this object is in the air for 9.26 s, at what velocity was it thrown?

d=vit +1/2at^2
d=v/t
a=vf-vi/t

## The Attempt at a Solution

x motion
---------
vi=vf because no acceleration
t=9.26 s
a=0

y motion
---------
a= -9.8
t=9.26
d=419.7m
vf=90.7m/s
vi=0

I arrived at vertical distance by the following:
90.7^2 = 0 + 2(-9.8)d
d=419.7
-----------------------
I arrived at final vertical velocity by the following:
-9.8=vf-0/9.26
vf=90.7m/s

I don't know if these are correct values or not, but even if they are...i don't know how they arrived at the final answer of 79.1m/s. Please help?

1irishman said:
d=vit +1/2at^2
This is the equation you want to use to analyze the vertical motion. Use this to solve for vi, which will be the vertical component of the initial velocity. (Then use some trig to find the total velocity, given that component.)

don't we have to know the value for d first though before we can solve for vi on the vertical? Is my value of d wrong above?

Presumably it starts and ends on the ground, so what is d (vertical position) after 9.26 s?

What's "d" (vertical displacement) if the object goes up and comes down to the same height?

zero?

so vi on vertical is 45.4m/s?

I have horizontal Rx=45.4/sin35=79.1, so the object was thrown initially at 79.1m/s right?

Yeah. You could have gotten this by noting that gravity decelerates objects at 9.8 m/s every second and that the ball took 9.26/2 s to reach maximum height, so the object was initially moving at 9.8*9.26/2 m/s.

yeah, but that would be too easy! Just kidding, thank you.

If i had used max height for this problem, could I have solved for vi then as well?

## 1. What is a projectile problem without air resistance?

A projectile problem without air resistance is a physics problem that involves calculating the motion of an object that is thrown or launched into the air without any air resistance or friction acting upon it.

## 2. How is the initial velocity of the object calculated in a projectile problem without air resistance?

The initial velocity of the object can be calculated using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity, and t is time.

## 3. How does the angle of launch affect the velocity in a projectile problem without air resistance?

The angle of launch affects the velocity in a projectile problem without air resistance because it determines the initial horizontal and vertical components of the velocity. A higher angle of launch will result in a higher initial vertical velocity, while a lower angle will result in a lower initial vertical velocity.

## 4. What is the maximum height reached by an object in a projectile problem without air resistance?

The maximum height reached by an object in a projectile problem without air resistance can be calculated using the equation h = (u*sinθ)^2/2g, where h is the maximum height, u is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity.

## 5. How can the range of the object be calculated in a projectile problem without air resistance?

The range of the object can be calculated using the equation R = u^2*sin2θ/g, where R is the range, u is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity. This equation assumes that the object is launched and lands at the same height.

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