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? Hello I need to use kinematic equations and the variables vfvitda

Answer 1

`154.6m`

Step-by-step explanation

First, let us make a sketch of the problem:

We want to find d from the diagram above.

To do this, we have to first find the time taken for the mattress to catch the human using the formula for vertical height below:

`y=y_0+v_(iy)t-(1)/(2)gt^2`

where y = ground level = 0 m

y0 = initial height of cannon = 0.75 m

viy = initial velocity in the y-direction

t = time taken

g = acceleration due to gravity

The components of velocity for the body are shown below:

This means that the velocity in the y-direction is given as:

`v_(iy)=v_i\sin \theta`

where vi = initial velocity, θ = angle of launch

Therefore, we have that:

`\begin{gathered} 0=0.75+40\sin 37(t)-(1)/(2)\cdot10\cdot t^2 \\ \Rightarrow0=0.75+24.07t-5t^2 \\ \Rightarrow5t^2-24.07t-0.75=0 \end{gathered}`

Solving for t using quadratic formula, we have that:

`t=4.84s;t=-0.031s`

Since time has to be positive, we have that:

`t=4.84s`

Now, we have to find the distance of the mattress from the cannon by applying the formula:

`d=v_(ix)t`

where vix = initial velocity in the x-direction.

We have that:

`v_(ix)=v_i\cos \theta`

Therefore, the distance is:

`\begin{gathered} d=v_i\cos \theta\cdot t \\ d=40\cdot\cos 37\cdot4.84 \\ d=154.6m \end{gathered}`