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A cannonball is shot (from ground level) with an initial horizontal velocity of 38.0 m/s and an initial vertical velocity of 26.0 m/s. The initial speed of the cannonball 46.04 m/s The initial angle θ of the cannonball with respect to the ground 34.38°

  1. What is the maximum height the cannonball goes above the ground? (m)
  2. How far from where it was shot will the cannonball land? (m)
  3. What is the speed of the cannonball 2.7 seconds after it was shot? (m/s)
  4. How high above the ground is the cannonball 2.7 seconds after it is shot? (m)
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What have you tried? What tools were you given to work with? –  Matthew Conroy Sep 12 '12 at 1:43
none.. the initial speed and angle I figured out and nothing ive tried is working. I have been trying this problem for like 2 hours –  Kayla Sep 12 '12 at 1:47
You can get help here, but you can't just dump a problem here and expect people to do it for you. Can you do any part of the problem? Have you seen any formulas that you think might be relevant? Can you follow the work on some similar problem you've seen? Please help us to help you. –  Gerry Myerson Sep 12 '12 at 1:49
Sorry, never used this site before. The most similar ive seen is how to get the inital speed and angle. I havent see equations for the rest. i know I need velocity and time equations. I just dont know what the equations are or how to set them up. –  Kayla Sep 12 '12 at 1:52
1. you only need vertical velocity. What is time when velocity is zero? What is height at that time. 2. What is time when displacement returns to original position? use that time with horizontal velocity to get distance. 3. Determine vertical speed given time. Then combine with horizontal speed. 4. Use time from (3) and use displacement formula. –  Tpofofn Sep 12 '12 at 1:53
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2 Answers 2

1) Use the formula v^2 = u^2 + 2as (v = final velocity, u = initial velocity, a = acceleration, and s = displacement)

We want final velocity in the y direction to be 0, because this is the case when the ball stops in the air, at maximum height.

Therefore in the y-direction,

0 = 26^2 + 2*-9.8*s Re-arrange to give, s = -26^2/(2*-9.8) = 34.5 m.

2) First we need to know the time the cannonball was in the air. We do this by solving the following equation in the vertical direction: v = u + at We set v = 0, to find the time the ball stops in the air at maximum height. This will give us half the total time the ball is in the air.

0 = 26 + -9.8t

Re-arrange to give,

t = -26/-9.8 = 2.65 seconds

Therefore total air time = 2*2.65 = 5.3 seconds.

Now we need to calculate the horizontal distance covered in 5.3 seconds. We use the formula for the horizontal variables: s = ut + 0.5*at^2 s = 38*5.3 [a = 0, because there is no acceleration in the horizontal direction] s = 201.4 m

Since there is no horizontal acceleration, horizontal speed = 38m/s for entire journey. We need vertical acceleration after 2.7 seconds. We use this equation in the vertical direction: v = u + at v = 26 + -9.8*2.7 v = -0.46 m/s

Total velocity = sqrt(x-component^2 + y-component^2) = sqrt(0.46^2 + 38^2) = 38m/s (approx) 4) We use the equation: s = ut + 0.5*at^2 [ for vertical direction ] s = 26*2.7 + 0.5*-9.8*2.7^2 s = 34.48 m

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I think it's terrible to tell someone to use a formula - it strengthens all the rubbish beliefs about mathematics that good teachers try to eradicate. –  Gerry Myerson Sep 13 '12 at 12:54
Can you please elaborate? How should I have explained this solution better and without using the formula? –  Mew Sep 17 '12 at 8:08
You could have derived the formula $v^2=u^2+2as$, or at least made some plausibility argument for it. –  Gerry Myerson Sep 17 '12 at 10:48
Ok, well I assumed that he would have had already been taught the formula before being given this question as homework. I don't think its necessary to derive a formula each time it is employed. Otherwise, it would defeat the whole point of formulas. –  Mew Sep 17 '12 at 11:19
Thank you Chris! And Gerry, if you can't say something nice, don't say nothing at all!!! –  user59700 Jan 26 '13 at 0:40
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a) $V_f^2 = V_0^2 + 2ad$
b) $V_f = V_0 + a(\frac{t}{2})$
$x = vt$
c) $V_x = V_{0x}$
$V_y = V_{0y} + at$
$V^2 = V_x^2 + V_y^2$
d) $d = V_{0y}t + \frac{1}{2}at^2$
Try and make something out of those.

Just as a hint, the vertical component of the velocity of the ball is 0 at its maximum height.

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The ultimate in mathematics-as-cookbook, only without a glossary. –  Gerry Myerson Sep 13 '12 at 12:55
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