Projectile Motion of Ball Problem

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Homework Statement



A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 9.20 m/s at an angle of 19.0° below the horizontal. It strikes the ground 4.00 s later.
(a) How far horizontally from the base of the building does the ball strike the ground?
(b) Find the height from which the ball was thrown.
(c) How long does it take the ball to reach a point 10.0 m below the level of launching?

Homework Equations



xf=xi+1/2(vxi +vxf)*t
xf=xi+vxi*t+1/2a*t2
vxf2= vxi2+2*a(xf-xi)
vf = vi + a*t
vxi= vi*cos([tex]\Theta[/tex])
vyi= vi*sin([tex]\Theta[/tex])
h= vi2s*in2 ([tex]\Theta[/tex])/(2g), g= gravity constant 9.8 m/s2, h= max height
R= vi2*sin(2[tex]\Theta[/tex])/g, R=max horizontal range

The Attempt at a Solution


I know how to do part (a). This is what I did
vxi= vi*cos[tex]\Theta[/tex]
vxi= 9.2*cos(19)= 8.70 m/s
xf=xf=xi+vxi*t+1/2a*t2
xf= 0+8.7*4 +0= 34.8 m

But for part (b) and (c), im stuck, and I cannot do part (c) w/o doing part (b). Heres my attempt.
What I did bascially is finding the time vyf = 0, which indicates the time of the maximum height. Then I used the yf equation to find the height of ball at that time. Then I just got stuck. Please help asap!
 
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Answers and Replies

  • #2
LowlyPion
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Homework Statement



A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 9.20 m/s at an angle of 19.0° below the horizontal. It strikes the ground 4.00 s later.
(a) How far horizontally from the base of the building does the ball strike the ground?
(b) Find the height from which the ball was thrown.
(c) How long does it take the ball to reach a point 10.0 m below the level of launching?

Homework Equations



The Attempt at a Solution


I know how to do part (a). This is what I did
vxi= vi*cos[tex]\vartheta[/tex]
vxi= 9.2*cos(19)= 8.70 m/s
xf=xf=xi+vxi*t+1/2a*t2
xf= 0+8.7*4 +0= 34.8 m

But for part (b) and (c), im stuck, and I cannot do part (c) w/o doing part (b). Heres my attempt.
What I did bascially is finding the time vyf = 0, which indicates the time of the maximum height. Then I used the yf equation to find the height of ball at that time. Then I just got stuck. Please help asap!

Think about the height equation Y = 1/2*a*t2

But you also have an initial velocity down of 9.2*Sinθ added to the 1/2(9.8)*(4)2

That yields your Total height Y = 9.2*Sinθ + 1/2(9.8)*(4)2

Now use the same equation but rather than plug in T=4 and solve for Y, plug in Y=10 and solve for T.
 
  • #3
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Think about the height equation Y = 1/2*a*t2

But you also have an initial velocity down of 9.2*Sinθ added to the 1/2(9.8)*(4)2

That yields your Total height Y = 9.2*Sinθ + 1/2(9.8)*(4)2

Now use the same equation but rather than plug in T=4 and solve for Y, plug in Y=10 and solve for T.

But I dont think that makes sense because you want the height of where you threw it from which was the upper story of the building, so u would have to find the time at which t equals same height as where you threw it from to find the initial height, but i have no clue on how to do that. Can you explain your process for part b again. Thanks
 
  • #4
LowlyPion
Homework Helper
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But I dont think that makes sense because you want the height of where you threw it from which was the upper story of the building, so u would have to find the time at which t equals same height as where you threw it from to find the initial height, but i have no clue on how to do that. Can you explain your process for part b again. Thanks

If you dropped it from that height of the building starting with 0 velocity, calling your height 0, and it takes 4 seconds to hit the ground how high is that?

You will find that this distance would be given by Y = 1/2*g*t2

Now if you threw it downward with an initial velocity will that make the distance longer if it still takes 4 seconds? Yes it will, by the value of the initial velocity times the time until you hit the ground. Hence that vertical velocity contribution to distance is the 9.2*Sine(19)*T where t=4 for this case.

Edit: (I see that I omitted the *T to the velocity term the first time. )
 
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  • #5
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Okay now i get it, why didnt I think of it. lol. I have a question though. Why dont we put -9.8 m/s^2 for the gravity constant instead of a positive 9.8, because the ball is falling down?
 
  • #6
LowlyPion
Homework Helper
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Okay now i get it, why didnt I think of it. lol. I have a question though. Why dont we put -9.8 m/s^2 for the gravity constant instead of a positive 9.8, because the ball is falling down?

So long as you are consistent with your signs, matching velocities and accelerations with what you call positive position, you can do it as you wish.
 

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