Problem 3: A ball is thrown straight up from ground level. It passes a 2-m-high

Question

Problem 3: A ball is thrown straight up from ground level. It passes a 2-m-high window. The bottom of the window is 7.5 m off the ground. The time that elapses from when the ball passes the bottom of the window on the way up, to when it passes the top of the window on the way back down is 1.3s What was the ball's initial speed, in meters per second? Numeric :A numeric value is expected and not an expression Problem 4: You throw a rock off of a cliff with a horizontal velocity, v. One second later you throw an identical rock with the same horizontal velocity, v. What is true about the distance between the two rocks beginning after you throw the second one? MultipleChoice 1) The distance between the rocks will decrease as time passes. l stay the same as time passes. 3) The distance between the rocks will increase as time passes. 4) It depends on the height of the cliff. 5) It cannot be determined.

Explanation / Answer

3. given

ball is thrown up from ground level

distance of bottom of the window from the ground = d = 7.5m

height of window , h = 2 m

let the initial velocity be v

then time taken to reach bottom of the window be t1

d = vt1 - 0.5g(t1)^2

velocity at this point = u

u = v - gt1

also, 2gd = v^2 - u^2

time taken to cross the bottom of the window and then top of the window and then come back tot he top of the window = t2 = 1.3 s

h = ut2 - 0.5gt^2

2 = u*1.3 - 0.5g*1.3^2

u = 7.914 m/s

hence 2*9.81*7.5 = v^2 - 7.914^2

v = 14.48 m/s

so initial velocity of the ball is 14.48 m/s

4. at t = 0, a rock is thrown off the cliff with horizontal velocity v

so at time t, its coordinates are

x = vt

y = -0.5gt^2

after 1 s, second rock is thrown

so coordinates are

x = v(t - 1)

y = -0.5g(t-1)^2

so distancne b/w rocks = sqroot((vt - vt + v)^2 + (0.5gt^2 - 0.5gt^2 -0.5g + gt )^2)

d = sqroot((v)^2 + (-0.5g + gt )^2)

so the distance b/w rocks increases with time

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