Projectile Motion Help, Please?

  • Thread starter bonita
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  • #1
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Well, I'm not sure how to end this problem. Is it correct how I approached it?

A basketball leaves a player's hands at a height of 2.1 m above the floor. The basket is 2.6 m above the floor. The player likes to shoot the ball at a 35* (degree) angle. If the shot is made from a horizontal distance of 12.0 m and must be accurate to +or- 0.22 m (horizontally), what is the range of initial speeds allowed to make the basket?

vx = v1cos35*
vy = v1sin35*

t = dx/v1cos35*
dx = 12.0 m
d = 0.5 m
d = v1t + 1/2gt^2

0.5 m = v1sin35* x (12.0m/v1cos35*) + 1/2(-9.8m/s^2)(12.0m/v1cos35*)^2
All cleaned up all that work and got v1 = 11 m/s.

dx = 12.0m + 0.22m = 12.22 m
dx = 12.0m - 0.22m = 11.78 m

t = dx/vx
t = 12.0m/11m/s = 1.1 seconds

dx = vxt
12.22m = vx(1.1seconds)
vx = 11.1 m/s
11.78m = vx(1.1seconds)
vx = 10.7 m/s

vx = v1cos35*
11.1 m/s = v1cos35* = 14 m/s
10.7 m/s = v1cos35* = 13 m/s

range of initial speeds = 13-14 m/s
 
Last edited:

Answers and Replies

  • #2
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Originally posted by bonita
vx = v1cos35*
vy = v1sin35*

t = dx/v1cos35*
dx = 12.0 m
d = 0.5 m
d = v1t + 1/2gt^2
I think the last one should read d =vyt + 1/2gt^2. It's OK since you used that in the next line:
0.5 m = v1sin35* x (12.0m/v1cos35*) + 1/2(-9.8m/s^2)(12.0m/v1cos35*)^2
All cleaned up all that work and got v1 = 11 m/s.
OK. I get v1 = 11.54 m/s.

Why not plug in dx = 11.78m resp. dx = 12.22m into the same formula?
This yields
v1(min) = 11.44 m/s
v1(max) = 11.63 m/s
 

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