<p>The magnitude of the velocity of a projectile when it is at its maximum heigh

Question

<p>The magnitude of the velocity of a projectile when it is at its maximum height above ground level is 19 m/s. <strong>(a)</strong> What is the magnitude of the velocity of the projectile 1.0 s before it achieves its maximum height? <strong>(b)</strong> What is the magnitude of the velocity of the projectile 1.0 s after it achieves its maximum height? If we take <em>x</em> = 0 and <em>y</em> = 0 to be at the point of maximum height and positive <em>x</em> to be in the direction of the velocity there, what are the <strong>(c)</strong><em>x</em> coordinate and <strong>(d)</strong><em>y</em> coordinate of the projectile 1.0 s before it reaches its maximum height and the <strong>(e)</strong><em>x</em> coordinate and <strong>(f)</strong><em>y</em> coordinate 1.0 s after it reaches its maximum height?</p>

Explanation / Answer

at its maximum height, velcoity vx = 19 m/s, vy = 0
(a) What is the magnitude of the velocity of the projectile t = 1.0 s before it achieves its maximum height?

vx = 19 m/s, vy = gt = 9.8 m/s, so v = (vx2 + vy2) = 21.4 m/s

(b) What is the magnitude of the velocity of the projectile 1.0 s after it achieves its maximum height?

vx = 19 m/s, vy = -gt = 9.8 m/s, so v = (vx2 + vy2) = 21.4 m/s

If we take x = 0 and y = 0 to be at the point of maximum height and positive x to be in the direction of the velocity there, what are the (c) x coordinate and (d) y coordinate of the projectile t = 1.0 s before it reaches its maximum height and the (e) x coordinate and (f) y coordinate 1.0 s after it reaches its maximum height?

c) x = -vxt = -19 m

d) y = -gt2/2 = -4.9 m

e) x = vxt = 19 m

f) y = -gt2/2 = -4.9 m

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