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anonymous
 one year ago
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.
anonymous
 one year ago
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.

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Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0Hi! Have you tried solving it on your own first?

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0Suppose, the rock is projected with a velocity V, then the two rectangular components we are interested in are the horizontal and vertical components.

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0Let 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
 one year ago
Best ResponseYou've already chosen the best response.0Finally, 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°

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Could you please write what the second equation of motion is without substituting in the values?

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0\(\sf s=ut+\frac{1}{2}\times a \times t^2\) s= distance u=initial velocity t= time a=acceleration.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So the acceleration value is the force of gravity?

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0Yes, 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
 one year ago
Best ResponseYou've already chosen the best response.0Note that there is no acceleration in horizontal direction hence we can use simply \(\sf Distance = speed \times Time\) for calculations in horizontal direction.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok, so \[V_{V} = \frac{ 9.75 + 1/2 * 9.8 * 16 }{ 4}\] ?
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