Spear and Target Problem : Spear released at an angle of 35 degrees and velocity

Question

Spear and Target Problem : Spear released at an angle of 35 degrees and velocity of 12 m/s. The center of the target and the spear release point are level. Target is 1m in diameter and target is 13m away from away.

1. Determine the distance the spear would
travel when it returns to the release height.
2. Does the spear hit high, low or directly on
the center of the target?
3. How many seconds does it take for the
spear to reach the target?
4. What is the vertical displacement of the
spear when it reachs the target?

Explanation / Answer

a. To find the horizontal displacement at 4.0 s : d = vivt + (0.5)at2 where viv = 30, t = 4.0 and a = 0 . Solve for d b. To find the horizontal component of velocity at 4.0 s : vfh = vih + at = 30 m/s + 0 = 30 m/s c. To find the vertical component of the velocity at 4.0 s : vfv = viv + at = 0 + (10)(4.0) = 40 m/s d. The resultant velocity at 4.0 s is v(302 + 402) = 50 m/s ø = tan-1(40/30) = 53º below the horizontal e. The vertical displacement at 4.0 s is: d = vivt + (0.5)at2 where viv = 0, t = 4.0 and a = 10 . Solve for d 2. First, find the components of the initial velocity: viv = 12sin40º = 7.71 m/s (+ is assigned up) vih = 12cos40º = 9.19 m/s

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