Physics, asked by tharishini285, 15 hours ago

A stone is thrown horizontally out to sea from the top of a vertical cliff and with a specd of 10 Oms. The cliff is 122,5m high. Find (1) the time it takes the stone to hit the water; (2) the speed of the stone and the direction in which it is travelling when it hits the water, (3) the horizontal distance from the bottom of the cliff at which the stone hits the water, If the stone hits a seagull, 3 seconds after it is thrown, find (4) its velocity when it hits the seagull; (5) how far the seagull was from the cliff when it was hit; (6) the height above the water at which the seagull was flying​

Answers

Answered by gondevedant93
0

Answer:

885438426554668556,555666

Answered by hotelcalifornia
0

Explanation 1

We have been given that the stone is released from a cliff of height 122.5m with initial velocity being 10m/s. For an object released from a height h, the velocity to be considered should be vertical only thus, the acceleration acting on the stone is acceleration due to gravity 'g'.

We have,

u_{y}=10m/s    ;  h=122.5m   ;  a=g=10m/s^{2}

Hence, time taken will be t

s=ut+\frac{1}{2}at^{2}

h=u_{y}t+\frac{1}{2}gt^{2}

122.5=(0)t+\frac{1}{2}(10)t^{2}

t^{2} =\frac{245}{10}

t=4.94sec

Hence, the stone takes 4.94 seconds to reach the sea.

Explanation 2

We know, in a projectile motion, the horizontal component of velocity is always constant but the vertical component of velocity becomes 0 at some point. Hence, the final velocity of the horizontal velocity will be 10 m/s only.

Now, the final velocity of the vertical component will be

v_{fv}=u_{iv}+at

We have, u_{iv}=10m/s   ; a=10m/s^{2}   ;  t=4.94s

Hence,

v_{fv}=10+10(4.94)

v_{fv}=59.4m/s

Therefore, the vertical component of velocity of the stone on reaching the sea will be 59.4 m/s.

Hence, the final velocity on reaching the sea will be

V=\sqrt{ (v_{fh} )^{2}+ (v_{fv} )^{2}

V=\sqrt{(10)^{2}+ (59.4)^{2} }

V=\sqrt{3628}

V=60m/s   (approx)

Hence the final velocity with which the stone reaches the sea is 60 m/s.

Explanation 3

It is given that the stone hits the seagull after 3 s of the start of the motion,

We have, for the horizontal motion,

u_{h}=10m/s   ; a=0m/s^{2}   ; t=3s  

Hence, distance traveled by the stone in 3 s is

s=ut+\frac{1}{2}at^{2}

s=10(3)+\frac{1}{2}(0)(3)^{2}  

s=30m

Hence, the stone hits the seagull at a distance of 30 m from the foot of the cliff.

Explanation 4

We know the stone hits the seagull after 3 s . Also, since the horizontal component of velocity remains constant always. Therefore, any change of velocity is purely the vertical velocity acting under acceleration of gravity.

We have,  U_{v}=10m/s   ; a=10m/s^{2}    ;  t=3s

v=u+at

V_{v}=10+10(3)

V_{v} =40m/s

Therefore, the velocity with which the stone hits the seagull

V=\sqrt{(V_{H} )^{2}+ (V_{V} )^{2} }

V=\sqrt{(10)^{2}+ (40)^{2} }

V=41.2m/s   (approx)

Explanation 5

We have that the initial velocity of the stone while throwing from the cliff as 10 m/s and it takes 3 s from the throw to hit the seagull. Hence, we have

u=10m/s   ;  t=3s   ; a=10m/s^{2}

s=ut+\frac{1}{2}at^{2}

s=10(3)+\frac{1}{2}(10)(3)^{2}

s=30+45

s=75m

Hence, the seagull is 75 m away from the cliff.

Explanation 6

We know, time taken by the stone to reach the water is 4.94 s and time taken to reach the seagull is 3 s.

Hence, after hitting the seagull, the remaining time that is 4.94-3=1.94s is the time taken by the stone to reach the ground.

Let the height at which seagull is flying be h,

We have s=h   ; t=1.94s   ; a=10m/s^{2}  ; u=0m/s

s=ut+\frac{1}{2}at^{2}

h=0(1.94)+\frac{1}{2}(10)(1.94)^{2}

h=18.8m

Hence, the seagull was flying at a height of 18.8 m from the surface of the water.

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