**Question 1**

The vertical height of the cliff is 86m

Acceleration due to gravity = 9.81m/s^{2}

Vertical component of velocity is zero

Hence applying newtons equation of motion

S= ut + 0.5gt^{2}

Here S = 86m and u = 0

Hence 86 = 0.5*9.81*t^{2}

Solving the above, we get t= 4.19 seconds

**Question 2**

Time taken to hit the ground = 4.19 seconds (Calculated from above)

Horizontal speed = 32m/s

Hence horizontal distance covered = 32*4.19 = 133.99 m or approximately 134 m.

**Question 3**

The vertical height of the cliff is 40m

Acceleration due to gravity = 9.81m/s^{2}

Vertical component of velocity is 25sin42 = 16.73m/s in the direction opposite to that of the ground

First we calculate the time required for the ball to reach the highest point

V= u + at

Hence 0 = -16.73 + 9.81*t_{1}

Hence t_{1} = 16.73/9.81 = 1.705 seconds

Distance travelled till the highest point be H

Hence 16.73^{2} = 2*9.81*H

Solving the above we get H = 14.265 m

Total height = 40 +14.265 = 54.265 m

Now applying s= ut + 0.5gt^{2}, we get

54.265 = 0*t_{2} + 0.5*9.81*t_{2}^{2}

Solving the above, we get t_{2} = 11.06 seconds

Hence total time of flight = 1.705 + 11.06 = 12.77 seconds

**Question 4**

Horizontal component of velocity = 25cos42 = 18.578m/s

Total time of flight = 12.77 seconds

Hence total horizontal distance from base of cliff = 18.578*12.77 = 237.23 m

**Question 5**

Speed of the rescue plane = 420 km/hr = 116.67 m/s

The velocity of the capsule would be only horizontal and the vertical component would be zero

The vertical height is given as 2263 m

Using the above data, let us calculate the time taken for the capsule to reach the ground

2263 = 0*t + 0.5*9.81*t^{2}

Solving the above we get t = 21.48 seconds

Horizontal distance covered by the capsule before touching the ground = 116.67*21.48 = 2505.93 m

Hence the required angle = tan^{-1}(2505.93/2263) = 47.9 degrees

**Question 6**

Height of the volcano = 3.2 km = 3200 m

Initial vertical component of the velocity of the block = 311sin 35 = 178.38 m/s

First we calculate the time required for the block to reach the highest point

V= u + at

Hence 0 = -178.38 + 9.81*t_{1}

Hence t_{1} = 178.38/9.81 = 18.18 seconds

Distance travelled till the highest point be H

Hence 178.38^{2} = 2*9.81*H

Solving the above we get H = 1621.83 m

Total height = 3200 + 1621.83= 4821.83 m

Now applying s= ut + 0.5gt^{2}, we get

4821.83 = 0*t_{2} + 0.5*9.81*t_{2}^{2}

Solving the above, we get t_{2} = 31.353 seconds

Hence total time of flight = 31.353 + 18.18= 49.54 seconds

**Question 7**

Horizontal component of velocity = 311cos35 = 254.76 m/s

Total time of flight = 49.54 seconds

Hence total horizontal distance from centre of volcano = 254.76*49.54 = 12,619.92 m

**Question 8**

Average speed = Total distance/Total time

Total distance = 76+63 = 139 km

Total time = 10+7 = 17 hours

Average speed = 139/17 = 8.18 km/hr

**Question 9**

Average velocity = Total displacement/Total time

Total displacement = √(76^{2}+63^{2}) = 98.716 km

Total time = 10+7 = 17 hours

Average velocity (magnitude) = 98.716/17 = 5.81 km/hr

**Question 10**

The direction of the average velocity would be the same as that of the displacement vector whose direction is north-west.