Question

A projectile is launched at ground level with an initial speed of 54.5 m/s at an angle of 35.0° above the horizontal. It strikes a target above the ground 2.80 seconds later. What are the x and y distances from where the projectile was launched to where it lands?

Answer 1

Answer: x=125m, y=48.308m

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which we have two components: x-component and y-component. Being their main equations to find the position as follows:

x-component:

$x=V_{o}cos\theta t$   (1)

Where:

$V_{o}=54.5m/s$ is the projectile's initial speed

$\theta=35\°$ is the angle

$t=2.80s$ is the time since the projectile is launched until it strikes the target

$x$  is the final horizontal position of the projectile (the value we want to find)

y-component:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$   (2)

Where:

$y_{o}=0$  is the initial height of the projectile (we are told it  was launched at ground level)

$y$  is the final height of the projectile (the value we want to find)

$g=9.8m/s^{2}$  is the acceleration due gravity

Having this clear, let's begin with x (1):

$x=(54.5m/s)cos(35\°)(2.8s)$   (3)

$x=125m$   (4)  This is the horizontal final position of the projectile

For y (2):

$y=0+(54.5m/s)sin(35\°)(2.8s)-\frac{(9.8m/s^{2})(2.8s)^{2}}{2}$   (5)

$y=48.308m$   (6)  This is the vertical final position of the projectile