A rock is launched from the ground with an initial velocity directed 56 degree a

Question

A rock is launched from the ground with an initial velocity directed 56 degree above the horizontal. (a) Assuming standard axes (x horizontal, y vertical) which quantity has a larger magnitude? Considering the interval from launch until the rock reaches its highest point, identify one quantity (by symbol or in a few words) that has a value of 0 in appropriate units. No explanation required or graded. Let the rock's speed at launch be 37 m/s. If it encounters no obstacle, how high will it rise? How high above the ground is the rock, 5.4 s after being launched? What is the rock's speed, 5.4 s after being launched? Is the rock is rising or falling, 5.4 s after being launched? Explain your choice in 1-2 sentence without additional math. At which of these three altitudes is the rock's speed the greatest? At which of these three altitudes is the rock's acceleration the greatest?

Explanation / Answer

a) Vix = vi cos56 = 0.56 vi

Viy = vi sin56 = 0.829 Vi

Viy is larger.

Ans. Viy

b) at highest point vertical component of velocity will be zero.

Vy = 0

c) Viy = 0.829 x 37 = 30.67 m/s

Vy^2 - Viy^2 = 2(a)(y)

0^2 - 30.67^2 = 2(-9.81)(y)

y = 48 m

d) y = viy*t + at^2 /2

y = (30.67 x 5.4) - (9.8 x 5.4^2 / 2)

y = 22.73 m

e) Vy = viy + at

     = 30.67 - 9.8x5.4

     = -22.25 m/s

f) velociy is negative and y is positive.

that means particle is above ground, and it it coming down.

so rock is falling.

g) as particl goes up it potential energy increases it KE decreases.

PE + KE = constant

so lesses the height, lesser the PE and more the KE

and greater the speed.

Ans . 1m

h) accleration is same everywhere .
(a = - g = - 9.8 m/s^2 )

same at all three altitudes.

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