Question

A boy throws a stone from the top of a cliff into the sea. When the stone leaves his hand it is horizontally travelling at a velocity of 10m/s. How long will it take the stone to reach the sea,a distance of 80 below?


(A) 1 sec
(C) 1.26 sec
(B) 1.5 sec
(D) None of these

Please solve this question in detail ​

Answer 1

Answer:

1.26s

Explanation:

Dist. covered in y direction is 8m

Initial velocity in y direction is 0m/s

Using y=ut+1/2at^2

8=0×t+1/2*10t^2

8=5t^2

=>t=1.26s

Answer 2

Concept:

The lateral and vertical movements are autonomous of one another. Newton's law of motion, which states that acceleration is equal to the acceleration due to gravity, provides the displacement. The stone's velocity profile in this scenario is zero.

Given:

From a cliff's edge, a boy throws a stone into the water. The stone is moving horizontally at a speed of 10 m/s when it exits his hand. How long will it take the stone to go the 80 feet to the sea?

(A) 1 sec

(B) 1.26 sec

(C) 1.5 sec

(D) None of these

Find:

Stated the provided statement's question, discover the acceptable answer.

Answer:

The answer is option (B). 1.26 sec

step-by-step explanation:

The stone is moving horizontally at a speed of $10m/s$ when it exits his hand.

$8m$ down is where the sea is.

Newton's law of motion states that

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

S stands for displacement, u for beginning velocity, t for time required to complete that displacement, and a for acceleration.

Gravitational acceleration is the force acting on the body. G is used to signify it.

Since $a=g,$

$s=y$ gives the displacement along the y-axis.

the formula for vertical displacement is $y=ut+\frac{1}{2} gt^{2}$

starting velocity $u=0$

$g=10m/s^{2}$ is the acceleration caused by gravity.

$y=0*t+\frac{1}{2} *10*t^{2}$

$y=8m\\8=0*t+\frac{1}{2} *10*t^{2} \\8=5t^{2}$

The stone's worth is determined by, and the time it takes to get there is t.

$t=\sqrt{\frac{8}{5} }$

$t=1.26sec$

Thus, the time required is $1.26 seconds$

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