A rock is projected from ground level. Later, 4.0s after being projected, the rock is observed to strike the top of a 9.75m tall fence that is a horizontal distance of 240ft from the point of projection. Determine the speed with which the rock was projected.

Question

A rock is projected from ground level.  Later, 4.0s after being projected, the rock is observed to strike the top of a 9.75m tall fence that is a horizontal distance of 240ft from the point of projection.  Determine the speed with which the rock was projected.

abhisar · Answer

Hi!

Have you tried solving it on your own first?

abhisar · Answer

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abhisar · Answer

Suppose, the rock is projected with a velocity V, then the two rectangular components we are interested in are the horizontal and vertical components.

anonymous · Answer

ok

abhisar · Answer

Let the vertical component be $\sf V_v$ and horizontal component be $\sf V_h$. We know that the rock travels a distance of 240 feets or 73.15 m in horizontal direction in 4 seconds. This is due to the horizontal component of its velocity.

$\sf \Rightarrow V_h=\frac{73.15}{4}$

Now, we also know that rock has traveled 9.75 m in vertically upward direction in 4 seconds. This must be due to the vertical component of its initial velocity. We can thus calculate the vertical velocity using second equation of motion.

$\Rightarrow \sf 9.75=V_v\times 4-\frac{1}{2}\times 9.8\times 4^2$

abhisar · Answer

Finally, after calculating the two components we can calculate its overall initial velocity by vectorally adding the two components. Remember that °since the components are rectangular, angle between them is 90°

abhisar · Answer

Getting it?

anonymous · Answer

Yeah, somewhat.

anonymous · Answer

Could you please write what the second equation of motion is without substituting in the values?

abhisar · Answer

$\sf s=ut+\frac{1}{2}\times a \times t^2$

s= distance
u=initial velocity
t= time
a=acceleration.

anonymous · Answer

So the acceleration value is the force of gravity?

abhisar · Answer

Yes, in this case acceleration on the rock is in vertically downward direction and I have opted to chose downwards as negative. So a here = $\sf -9.8 m/s^2$

abhisar · Answer

Note that there is no acceleration in horizontal direction hence we can use simply $\sf Distance = speed \times Time$ for calculations in horizontal direction.

anonymous · Answer

Ok, so $$V_{V} = \frac{ 9.75 + 1/2 * 9.8 * 16 }{ 4}$$ 
?

anonymous · Answer

@Abhisar

abhisar · Answer

Yes.