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Ideal Projectile Motion , horizontal and vertical components?

  • Thread starter djester555
  • Start date
10
0
1. Homework Statement

You throw a ball toward a wall with a speed of 32 m/s at an angle of 40.0° above the horizontal directly toward a wall. The wall is 22.0 m from the release point of the ball.


(a) How far above the release point does the ball hit the wall?


(b) What are the horizontal and vertical components of its velocity as it hits the wall?


3. The Attempt at a Solution

a).
(Ux)* t = Sx
(32cos40)* t = 22
t = 22/(32cos40)

Sy = (Uy)t+0.5(Ay)t^2
= {(32sin40)*[22/(32cos40)]
+0.5*(-9.8)*[22/(32cos40)]}m
= 12.16m i know this is wrong


b)b).
Vx= 32cos40
= 24.513m/s this is correct


(Vy)^2 = (Uy)^2+2(Ay)*(Sy)
= (32sin40)^2+2*(-9.8)*(-12.16)
Vy = 25.72m/s this is wrong

having problems with vertical component and part A to the question
 

Andrew Mason

Science Advisor
Homework Helper
7,521
301
You correctly found the time it hits the wall: t = d/vx = 22/32 cos40

You used the correct formula for height: h = h0 + vyt - .5gt^2 but you forgot to square t!

AM
 
10
0
a).
(Ux)* t = Sx
(32cos40)* t = 22
t = 22/(32cos40)

Sy = (Uy)t+0.5(Ay)t^2
= {(32sin40)*[22/(32cos40)]
+0.5*(-9.8)*[22/(32cos40)]^2}m
-4.9*1.0626
= -5.20679
this still seems to be wrong
 

Andrew Mason

Science Advisor
Homework Helper
7,521
301
a).
(Ux)* t = Sx
(32cos40)* t = 22
t = 22/(32cos40)

Sy = (Uy)t+0.5(Ay)t^2
= {(32sin40)*[22/(32cos40)]
+0.5*(-9.8)*[22/(32cos40)]^2}m
-4.9*1.0626
= -5.20679
this still seems to be wrong
I get 14.5 m:

t = 22/32cos40 = 22/24.5 = .90 sec.

h = vyt - .5gt^2 where vy = 32sin40= 20.5 m/s
= 20.5(.90) - .5 x 9.8(.9^2)
= 18.5 - 4.0
= 14.5 m

AM
 

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