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chrisfnet
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Homework Statement
A ball is fired from a height of 3.0m above the ground, with a speed of 13m/s and an angle of 22 degrees below horizontal. (a) At what time does the ball hit the ground? (b) What is the impact velocity?
Homework Equations
vx = v0x + axt
vy = v0y + ayt
∆y= v0∙t+1/2a∙t^2
vx = v0x + axt
vy = v0y + ayt
The Attempt at a Solution
The part I'm really not sure about is what angle I should derive the components from. Any input would be appreciated!
ay = -9.8m/s2
ax = 0m/s2
v0 = 13m/s
∆y = -3.0m
v0x = ?
v0y = ?
t = ?
∆x = ?
v0x = cos 22∙(13m/s) = 12.05m/s
v0y = sin 22∙(13m/s) = 4.87m/s
Solve for 't'
∆y= v0∙t+1/2a∙t^2
-3.0m = -13m/s ∙ t + 1/2 ∙ (-9.8m/s2) ∙ (t2)
t = 0.214s
I made the velocity negative, indicating its downward path. Corrections are in red.
Solve for impact velocity.
vx = v0x + axt
vx = 12.05m/s + (0m/s2) ∙ (0.214s)
vx = 12.05m/s
vy = v0y + ayt
vy = 4.87m/s + (-9.8m/s2) ∙ (0.214s)
vy = 2.77m/s
v = 12.05m/s(x hat) + 2.77m/s(y hat)
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