We’re asked to find the total time the paintball is in the air. This is equal to

Question

We’re asked to find the total time the paintball is in the air. This is equal to the time it would take the paintball to fall vertically from its initial height to the ground. In each case, the vertical position is given as a function of time by the equation, y=y0+v0yt12gt2. In this problem, y0=1.50m, y=0, and v0y=0, so that equation becomes simply

0=y012gt2

We need to find the time t. Solving for t and substituting numerical values, we obtain

t=2y0g=2(1.50m)9.80m/s2=0.553s

Part (b): Now that we know the time t of the ball’s flight through the air, we can find the range—that is, the horizontal distance x it travels during time t. We use the equation, x=x0+v0xt. Setting x0=0, we find

x=x0+v0xt=0+(75.0m/s)(0.553s)=41.5m

If air resistance is ignored, what initial speed is required for a range of 19m?

Explanation / Answer

Here ,

y0 = 1.50 m

part a)

y = y0 + vo * t - 0.5 * gt^2

0 = 1.50 - 0.5 * 9.8 * t^2

solving for t

t = 0.553 s

part b)

let the initial speed is v0

19 = v0x * 0.553

solving for vox

vox = 34.3 m/s

the initial speed of ball is 34.3 m/s

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