projectile motion (1 Viewer)

Aldehyde · Oct 11, 2009

1) Show for a projectile with a range R, and time of flight T, that:

tan (theta) = gT^2 / 2R

2) An archer shoots an arrow at a horizontal speed of of 160ms directly at target 30m away. The arrow is aimed directly at the target centre.

a) How far below the center will the target hit?
b) If the arrow is to hit the target center, at what angle to the horizontal should it be?

untouchablecuz · Oct 11, 2009

$x=Vtcos\theta$
When t=T, x=R
$R=VTcos\theta\Rightarrow VT=\frac{R}{cos\theta} (1)$
$y=-\frac{gt^2}{2}+Vtsin\theta$
When t=T, y=0
$0=-\frac{gT^2}{2}+VTsin\theta\Rightarrow \frac{gT^2}{2}=VTsin\theta (2)$
Sub (1) in (2)
$\frac{gT^2}{2}=sin\theta(\frac{R}{cos\theta})\Rightarrow tan\theta=\frac{gT^2}{2R}$

Aldehyde · Oct 12, 2009

bump...I need help with Q2 please.

tommykins · Oct 12, 2009

projectile motion (1 Viewer)

Aldehyde

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untouchablecuz

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Aldehyde

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tommykins

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