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Unformatted text preview: t’d
Solution, ∆dy = v1y ∆ t + ½ a ∆ t²
1y 20.0 m = 0 + 4.9t²
– Therefore, t = 2.019s
– Since time is equal in each direction,
Since S:∆dx = v1xt = 18(2.019) = 36.3 m P: Therefore, the arrow travels 36.3 m before hitting the
ground. Objects Launched at Angles
Objects Harder stuff. Now we need to combine
some We can always take the “launch” velocity of
a projectile and break it down from v1 to
component form: v1x and v1y
Projectiles A few things are important to remember here:
– The higher the launch angle, the greater the max height
– As a projectile reaches its maximum height, the vertical
velocity (vy) is 0m/s
– If the starting and finishing heights are the same (e.g. a
golf ball is hit from the ground and ends up on the
ground again) then the total time for the whole situation
is TWICE the time taken to reach max height.
– We can use a shortcut formula for max height h = v1y²
2g Everyone now try p. 82 #1...
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