A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.19 s s

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A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.19 s s

later. You may ignore air resistance. Part A Part complete If the height of the building is 21.0 m m , what must the initial speed be of the first ball if both are to hit the ground at the same time? v v = 9.53 m/s m/s SubmitPrevious Answers Correct Part B Part complete Consider the same situation, but now let the initial speed v 0 v0 of the first ball be given and treat the height h h of the building as an unknown. What must the height of the building be for both balls to reach the ground at the same time for v 0 v0v_0 = 8.70 m/s m/s . h h = 6.51 m m SubmitPrevious Answers CorrectPart C Part complete If v 0 v0 is greater than some value v max vmax , a value of h h does not exist that allows both balls to hit the ground at the same time. Solve for v max vmax . v max vmax = 11.7 m/s m/s SubmitPrevious Answers Correct Part D Part complete If v 0 v0 is less than some value v min vmin , a value of h h does not exist that allows both balls to hit the ground at the same time. Solve for v min vmin . v min vmin = 5.83 m/s m/s SubmitPrevious Answers Correct Provide Feedback Next

Physics

1 answer:

ziro4ka [17] · Answer 1 · 2022-06-25T00:29:55+03:00

A) Initial velocity of ball 1: 9.53 m/s upward

B) Height of the building: 6.48 m

C) Maximum velocity: 11.7 m/s

D) Minimum velocity: 5.83 m/s

Explanation:

A)

The y-position of the 1st ball at time t is given by the equation for free fall motion:

$y_1 = h + v_0 t - \frac{1}{2}gt^2$ (1)

where

h = 21.0 m is the initial height of the ball, the height of the building

$v_0$ is the initial velocity of the ball, upward

$g=9.8 m/s^2$ is the acceleration of gravity

The y-position of the 2nd ball instead, dropped from the roof 1.19 s later, is given by

$y_2 = h-\frac{1}{2}g(t-1.19)^2$

where

h = 21.0 m is the initial height of the ball, the height of the building

t' = 1.19 s is the delay in time of the 2nd ball (we can verify that at t = 1.19 s, then $y_2=h$ , so the ball is still on the roof

The 2nd ball reaches the ground when $y_2=0$ , so:

$0=h-\frac{1}{2}g(t-1.19)^2\\0=(21.0)-4.9(t^2-2.38t+1.42)\\4.9t^2-11.66t-14.04=0$

Which has two solutions:

t = -0.88 s (negative, we discard it)

t = 3.26 s (this is our solution)

The 1st ball reaches the ground at the same time, so we can substitute t = 3.26 s into eq.(1) and $y_1=0$ , so we find the initial velocity:

$0=h+v_0 t -\frac{1}{2}gt^2\\v_0 = \frac{1}{2}gt-\frac{h}{t}=\frac{1}{2}(9.8)(3.26)-\frac{21.0}{3.26}=9.53 m/s$

B)

In this case, the height of the building h is unknown, while the initial velocity of ball 1 is known:

$v_0 = 8.70 m/s$

When the two balls reach the ground at the same time, there position is the same, so we can write:

$y_1=y_2\\h+v_0 t - \frac{1}{2}gt^2 = h-\frac{1}{2}g(t-1.19)^2$

Solving the equation, we find:

$v_0t=1.19gt-\frac{1}{2}g(1.19)^2\\t=\frac{0.5g(1.19)^2}{1.19g-v_0}=2.34 s$

This is the time at which both balls reache the ground; and substituting into the eq. of ball 2, we find the height of the building:

$0=h-\frac{1}{2}g(t-1.19)^2\\h=0.5g(t-1.19)^2=0.5(9.8)(2.34-1.19)^2=6.48 m$

C)

If $v_0$ is greater than some value $v_{max}$ , then there is no value of h such that the two balls hit the ground at the same time. This situation occurs when the demoninator of the formula found in part b:

$t=\frac{0.5g(1.19)^2}{1.19g-v_0}$

becomes negative: in that case, the time becomes negative, so no solution is possible.

The denominator becomes negative when

$1.19g-v_0 < 0$

Therefore when

$v_0>1.19g=(1.19)(9.8)=11.7 m/s$

So, if the initial velocity of ball 1 is greater than 11.7 m/s, the two balls cannot reach the ground at the same time.

D)

There is also another condition that must be true in order for the two balls to reach the ground at the same time: the time at which ball 1 reaches the ground must be larger than 1.19 s (because ball 2 starts its motion after 1.19 s). This means that the following condition must be true

$t=\frac{0.5g(1.19)^2}{1.19g-v_0}>1.19$