How long it takes for the stone to fall to the water? Answer in units of s. With what speed does the stone strike the water? Answer in units of m/s.      A person standing at the edge of s...

How long it takes for the stone to fall to the water? Answer in units of s. With what speed does the stone strike the water? Answer in units of m/s.

 

 

 

A person standing at the edge of s seaside cliff kicks a stone over the edge with a speed of 14 m/s. The cliff is 62 m above the water's surface, and the acceleration of gravity is 9.81 m/s^2.

 

 

 

Asked on by jayjay00

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krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

The stone has a movement which can be resolved in a horizontal and vertical movement.

The vertical component of movement consists a continuously accelerating movement starting at 0 and accelerating at 9.81 m/s^2.

The stone hits the water surface when it has travelled 62 m vertically.

We know: t = (2s/a)^(1/2)

where t = time of travel, s = distance travelled, and a = acceleration.

Therefore: t = (2*62/9.81)^(1/2) = 3.5553 s (approximately)

Vertical component of speed of the stone then it hits the water

= a*t = 9.81*3.5553 = 34.8774 m/s

Speed of the stone when it hits the water =

= [(horizontal component of speed)^2 + (vertical component of speed)^2]^1/2

= (34.8774^2 + 14^2)^1/2 = 37.5824 m/s (approximately)

Answer:

The stone takes 3.5553 s to fall to water.

The stone strikes the water with sped of 37.5824 m/s.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

We assume the speed imparted to the stone is horizontal direction andso 14m/s is the horizontal velocity of the stone.The depth of the cliff is 62m.

To find the time t the stone reaches the water surface which is at the depth of 62m.

The horizotal component of velocity of the stone at any time t = 14m/s is constant and not affected by g as long as it is in motion.

The vertical component of velocity u is downwards = 0 initially and at the time of touching the water surface = g*t

Therefore the displacement s from the cliff down the stone falls is given by the equation:

s = (1/2)g t^2. But s= 62 here.

So, t= (2s/g)^(1/2)=(2*62/9.82)^(1/2)=3.5553s

Therefore the vertical component of velocityof the stone when it reaches water surface= gt =9.81*3.5553=34.8775m/s

The constant horizontal component of the velocity of the stone = 14m/s

Therefore the resultant velocity of the stone = {vertical compnent vel^2+horintal component of vel ^2}^(1/2) = (34.8775^2+14^2)^(1/2) =37.5824m/s

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