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projectiles questionn

How do you do the last question?
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Reply 1
You're looking for values of t for which vertical displacement is 5m (since it starts 1m above the ground). The setup implies that it just passes the wall when it's coming down, so take the greater value of t. Once you find that, use the horizontal velocity component to calculate distance with distance = velocity * time
Reply 2
Original post by Sinnoh
You're looking for values of t for which vertical displacement is 5m (since it starts 1m above the ground). The setup implies that it just passes the wall when it's coming down, so take the greater value of t. Once you find that, use the horizontal velocity component to calculate distance with distance = velocity * time

Thanks but for the vertical displacement bit in the question it says "She throws a stone as before and it just passes over the wall" Why are we assuming that the highest vertical displacement reached is when the stone just passes over the wall?
Reply 3
Original post by h26
Thanks but for the vertical displacement bit in the question it says "She throws a stone as before and it just passes over the wall" Why are we assuming that the highest vertical displacement reached is when the stone just passes over the wall?


It's not, so you're going to have an equation with two solutions. It reaches its highest point somewhere before the wall and just skims it on its way down.
the trajectory is the same size and shape as before, it just starts further to the left.

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