Question

Kinematic equation 1: Vi=V; + gt Kinematic equation 2: Y=V;.t + 12 gt? Kinematic equation 3: 2aY=V7-V21 22. A stone is thrown in a vertically upward direction with a velocity of Vi= 5 m/s. If the acceleration(g) of the stone during its motion is 10 m/s2 in the downward direction, A) what will be the height (Y) attained by the stone? B) how much time (t) will it take to reach there? Take g=-10 m/s2. [2] - [Hint: stone thrown upwards will stop at some time so final velocity will be ZERO] 23. Suppose that you drop (freefall) a marble from the top of Burj Khalifa building in Dubai, Which is about Y=830 m tall. If you ignore the air resistance, A) How long (t) will it take for the marble to hit the ground? B) What will be its final velocity (V just before it hits the ground? Take g=-9.8 m/s2. [2]

Answer 1

22) Given the initial velocity of the stone is Vi = 5m/s and acceleration g = -10m/s2. At the maximum height reached by the stone, its final velocity Vf = 0.

(A) The height Y attained by the stone can be found by using the kinematic equation,

2gY = VP-V,

21-10) Y = 02 - 54

25 Y = = 1.25m 20

So the height attained by the stone is 1.25m.

(B) The time taken t can be found using the kinematic equation,

$V_{f}=V_{i}+gt$

$0=5+(-10)t$

$10t=5$

$t=\frac{5}{10}=0.5s$

So the time taken to reach the height is 0.5s.

23) Since the ball is drropped Vi = 0. Given the height of the building Y = 830m and g = -9.8m/s2

(A) The time taken for the marble to hit the ground can be found using the kinematic equation,

$Y = V_{i}t+\frac{1}{2}gt^{2}$

Taking downward motion to be negative,

$-830 = 0\times t+\frac{1}{2}\times -9.8\times t^{2}$

$t^{2}=\frac{830}{4.9}=169.39$

$t=13.01s$

So the time taken by the marble to hit the ground is 13.01s.

(B) The final velocity can be found using the equation.

2gY = VP-V,

$2\times -9.8\times -830=V{_{f}}^{^{2}}-0^{2}$

$V{_{f}}^{2}=16268$

$V_{f}=127.55m/s$

So the final velocity of the marble before it hits the ground is 127.55m/s.