Solving Projectile Motion: Horizontal & Vertical Distances

[kitle545]

Homework Statement

A projectile is fired at an angle of 60 degrees above the horizontal with an initial speed of 300 m/s. Calculate A) the horizontal distance traveled and B) the vertical height attained in the first 6 s.

Homework Equations

The Attempt at a Solution

I did 300cos60=150 so Vi(x)=150 and 300sin60=259.8 so Vi(y)=259.8 but after that it seems like i don't have enough information to continue.

 

[Rake-MC]

you need to use constant acceleration formulae

we know:
Ux = 150ms^-1
Uy = 259.8ms^-1
a = -9.81ms^-2

First you can calculate how long it's in the air for, we know that its trajectory is symmetrical about the peak, and at the peak it's y velocity is 0ms^-1.

$$v = u + at$$


$$t = \frac{v-u}{a}$$

= $$\frac{0 - 259.8}{-9.81} = 26.48s$$

For it to hit the ground again, it must have been in the air twice that time = 53s.

now you know for the x plane, a, t and u
find s (displacement)

$$s = ut + \frac{1}{2}at^2$$

hint: half of that equation dissapears when you sub in the values.

You can do the rest yourself!

 

[baba89]

I would like to clarify that there are a few assumptions that need to be made in order to solve this problem. Firstly, we need to assume that the projectile is fired in a vacuum, meaning that there is no air resistance. Secondly, we can assume that the acceleration due to gravity is constant at 9.8 m/s^2.

Using these assumptions, we can solve for the horizontal distance traveled and vertical height attained in the following ways:

A) The horizontal distance traveled can be calculated using the formula d = v*t, where d is the distance, v is the initial velocity, and t is the time. In this case, we can use the initial horizontal velocity of 150 m/s and the time of 6 seconds to calculate the horizontal distance traveled. Therefore, d = 150 * 6 = 900 meters.

B) To calculate the vertical height attained, we can use the formula h = v*t + (1/2)*a*t^2, where h is the height, v is the initial vertical velocity, a is the acceleration due to gravity, and t is the time. In this case, we can use the initial vertical velocity of 259.8 m/s and the time of 6 seconds to calculate the vertical height attained. Therefore, h = 259.8 * 6 + (1/2)*9.8*(6^2) = 934.8 meters.

In conclusion, the horizontal distance traveled is 900 meters and the vertical height attained is 934.8 meters in the first 6 seconds. It is important to note that these calculations are based on the assumptions made and may differ in real-life scenarios where air resistance and non-constant acceleration due to gravity are present.