Homework Statement
A basketball is thrown horizontally with an initial speed of 4.20 m/s. A straight line drawn from the release point to the landing point makes an angle of 30.0° with the horizontal. What was the release height?
Vox=4.20 m/s=Vx Ax=0
Voy=0 m/s Ay=-9.81 m/s^2
Theta=180*-30.0*...
okay, for question 1:
0^2=(130m/s)^2+2a(0-22)
-16900=-44a
a=384.90m/s^2
and for question 2:
(change in)x=[0^2-(16m/s)^2]/[2(-3.2)m/s^2]=40m
55m-40m=15m
[15m]/[16m/s]=0.9375s=~0.94s
thanks for the kick in the right direction guys! great website!
Quick "Equations of Motion" questions (2). Rxn time; deceleration
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