**A rescue plane is flying horizontally with a speed of 50 m s ^{-1} and at an altitude of 200 m above the sea when it drops a warning flare.**

**Neglecting air resistance and assuming that the plane does not change its course, speed or altitude, how far from the plane is the flare when it hits the water?**

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The flare will be released with the same horizontal velocity of 50 m s

^{-1}. Since there is no acceleration in the horizontal direction, the flare will be directly below the position of the plane when it hits the water

Answer: 200 m

**A boy throws a ball vertically upwards. The ball falls back into his hand 2.0 seconds later. Neglecting air resistance, what is the speed at which the ball leaves his hand?**

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Taking upward motion as positive,

s = ut + ½ at

^{2}

0 = u(2) + ½ (-9.81)(2

^{2})

u = ½ (9.81)(2)

u = 9.81 m s

^{-1}

**A ball was kicked over a 8.0m wall as shown below with a velocity u at an angle of 50° above the horizontal. At the highest point of the trajectory, the ball managed to just go over the wall. It landed into a depression 2.0 m deep.**

**a) Express the initial horizontal velocity and initial vertical velocity in terms of u.**

**b) Calculate the initial speed of the ball**

**c) Calculate the vertical velocity of the ball v _{y} just before it hits the depression.**

**d) Hence, or otherwise, calculate the velocity of the ball v just before it hits the depression, and the angle it makes below the horizontal.**

**e) Calculate the time of flight**

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a) u

_{x}= u cos 50°

u

_{y}= u sin 50°b) Taking upwards and rightwards as positive, consider the ball at its highest point of its trajectoryv

_{y}

^{2}– u

_{y}

^{2}+ 2a

_{y}s

_{y}

0 = (u sin 50°)

^{2}+ 2(-9.81)(8.0)

u = 16.355

u = 16 m s

^{-1}

c)

v

_{y}

^{2}– u

_{y}

^{2}+ 2a

_{y}s

_{y}

v

_{y}

^{2}= (16.355 x sin 50°)

^{2}+ 2(-9.81)(-2.0)

v

_{y}

^{2}= 196.2

v

_{y}= -14.007

v

_{y}= -14 m s

^{-1}

d) v^{2} = v_{x}^{2} + v_{y}^{2}

v^{2} = (16.355 cos 50°) + (-14.007)^{2}

v = 17.513

v = 17 m s^{-1}

tan θ = (v_{y} / v_{x})

tan θ = (14.007 / 16.355 cos 50°)

θ = 53.111°

θ = 53°

The velocity of the ball is 17 m s^{-1} at an angle of 53° below the horizontal.

e) v_{y} = u_{y} + a_{y}t

-14.007 = (16.335 x sin 50°) + (-9.81)

t = 2.7049

t = 2.7 s

**The acceleration-time graph of an object moving in a straight line is as shown. The object started its motion from rest.**

**At which point is the body moving with the largest speed?**

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Area under acceleration-time graph = change in velocity of the object.At point 2, the area under the graph is the largest.

Answer: 2

Thanks a lot for sharing simple notes and questions, really helpful for self studying thanks a lot. : )