In an effort to test the laws of physics, you throw amedium-sized marble vertica

Question

In an effort to test the laws of physics, you throw amedium-sized marble vertically upwards into the air with an initialvelocity of 17.4 m/s. Your hand is 1.3 m above the ground when yourelease the marble. Use upwards as the positive direction. 1. What is the highest point above your hand which the marblewill reach? 2. From the time the marble leaves your hand to the time ithits the ground, what is the marble's total displacement? 3. What is the marble's velocity as it passes your hand on theway down? 4. What is the marble's velocity at the instant it hits theground? 5. The situation described in this problem is an applicationof which law? In an effort to test the laws of physics, you throw amedium-sized marble vertically upwards into the air with an initialvelocity of 17.4 m/s. Your hand is 1.3 m above the ground when yourelease the marble. Use upwards as the positive direction. 1. What is the highest point above your hand which the marblewill reach? 2. From the time the marble leaves your hand to the time ithits the ground, what is the marble's total displacement? 3. What is the marble's velocity as it passes your hand on theway down? 4. What is the marble's velocity at the instant it hits theground? 5. The situation described in this problem is an applicationof which law?

Explanation / Answer

vf = vi - gt vi = gt vi/g = t = (17.4/9.81) = 1.77 seconds to reach it's max height hmax = *t = (17.4/2)*1.77 = 15.4 meters = average! Okay. This 15.4 meters is measured using the initial height = 0,when it is actually 1.3 This answer satisfies part 1. But if we want to know how high it is above the ground, it is15.4+1.3 = 16.7 m 2. Displacement is the difference from start position to endposition. If we consider +1.3 as the starting position, 16.7 is themax height, then 0 is the ground. 16.7 meters. This could be wrong,depends on how you read the question (displacement questions still"get" me). 3. Just find the speed of the marble using 0 as the hand distanceand 15.4 as the height above it. vf^2 = 2*a*d vf = sqrt(2*g*15.4) (g is positive now). = 17.4 m/s. What weexpected. 4. Use the same equation, but change 15.4 to 16.7 vf^2 = 2*a*d vf = sqrt(2*9.81*16.7) = 18.1 m/s 5. Which law... Newton's laws? I'd say first law. Could also beconstrued as a conservation of energy verification... mgh=1/2mv^2.

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