Projectile Motion/Trajectories

  • Thread starter kobylorat
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Homework Statement



I've attached the problem. The answer key says the answer is (a), but I've reasoned it out to be (c). I'm not sure if the answer key is wrong, so can someone check my work? Thanks!

Homework Equations




The Attempt at a Solution



I've found that the initial vertical and horizontal speeds to be both 4√2 m/s using 8sin45 and 8cos45 respectively.
Then, dealing with only the vertical motion, I've found the time the water is in the air for by using the kinematic equation x = xi + vi*t + 0.5at^2.
Plugging values into the equation, I get: 1.5 = 4√2*t - 4.9t^2.
Solving for t, I get t=0.41269s and t=0.74177s. But, I don't use t=0.41269s because that is the time it takes for the ball to reach 1.5m as it is going up.
So, I take t=0.74177s and multiply it by the horizontal velocity, 4√2 m/s to obtain d=4.196084m, which means x is 2.19m.
 

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Answers and Replies

  • #2
1,065
10

Homework Statement



I've attached the problem. The answer key says the answer is (a), but I've reasoned it out to be (c). I'm not sure if the answer key is wrong, so can someone check my work? Thanks!

Homework Equations




The Attempt at a Solution



I've found that the initial vertical and horizontal speeds to be both 4√2 m/s using 8sin45 and 8cos45 respectively.
Then, dealing with only the vertical motion, I've found the time the water is in the air for by using the kinematic equation x = xi + vi*t + 0.5at^2.
Plugging values into the equation, I get: 1.5 = 4√2*t - 4.9t^2.
Solving for t, I get t=0.41269s and t=0.74177s. But, I don't use t=0.41269s because that is the time it takes for the ball to reach 1.5m as it is going up.
So, I take t=0.74177s and multiply it by the horizontal velocity, 4√2 m/s to obtain d=4.196084m, which means x is 2.19m.

You should find the height of water jet when it had travelled horizontally for 2m.
 

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