Question

The SPH3U class designs a cannon able to shoot a human being out of it. . If the human is launched at a velocity of 40 m/s, 37° from the ground, how far from the cannon should you place a mattress to catch the human if the muzzle of the cannon is 0.75 m from the ground?

Answer 1

Given:

Velocity = 40 m/s

Angle = 37°

Let's find the distance from the cnnon where you should place a matress to catch the human if the muzzle is 0.75 meters from the ground.

Let's first draw the projectile motion diagram:

To find the distance, first apply the formula to solve for the time taken for the human to reach the matress.

`y=y_o+v_(oy)t-(1)/(2)gt^2`

Where:

Height from the ground to the muzzle = yo = 0.75

y = 0

Velocity = vo = 40 m/s

initial vertical velocity = voy = vo sinθ = 40 sin37

We have:

`v_(oy)=v_o\sin \theta=40\sin 37=24.07\text{ m/s}`

g is the acceleration due to gravity = 9.8 m/s²

Now, let's solve for the time t:

`\begin{gathered} y=y_o+v_(oy)t-(1)/(2)gt^2 \\ \\ 0=0.75+24.07t-(1)/(2)*9.8* t^2 \\ \\ t=4.94\text{ seconds} \end{gathered}`

To sove for the horizontal distance, x, apply the formula:

`\begin{gathered} x=v_(ox)t \\ \\ \text{Where:} \\ v_(ox)=v_o\cos \theta=40\cos 37=31.94\text{ m/s} \end{gathered}`

Hence, we have:

`\begin{gathered} x=31.94*4.94 \\ \\ x=157.8\text{ m} \end{gathered}`

Therefore, the matress should be placed at 157.8 meters from the cannon in order to catch the human.

ANSWER:

157.8 meters