Physics, 08.10.2020 17:01 diamondgodbee123

A boy stands on a diving board and tosses a stone into a swimming pool. The stone is thrown from a height of m above the water surface with a velocity of m/s at an angle of above the horizontal. As the stone strikes the water surface, it immediately slows down to exactly half the speed it had when it struck the water and maintains that speed while in the water. After the stone enters the water, it moves in a straight line in the direction of the velocity it had when it struck the water. If the pool is m deep, how much time elapses between when the stone is thrown and when it strikes the bottom of the pool?

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Answer from: jayy6863

Complete Question

A boy stands on a diving board and tosses a stone into a swimming pool. The stone is thrown from a height of 2.50 m above the water surface with a velocity of 4.00 m/s at an angle of 60.0 degrees above the horizontal. As the stone strikes the water surface, it immediately slows down exactly half the speed it had when it struck the water and maintains that speed while in the water. After the stone enters the water, it moves in a straight line in the direction of the velocity it had when it struck the water. If the pool is 3.00 m deep, how much time elapses between the stone is thrown and when it strikes the bottom of the pool?

The value is $t_t = 1.9182 \ s$

Explanation:

From the question we are told that

The height of the throw is $h = 2.50 \ m$

The velocity is $v = 4.00 \ m/s$

The angle is $\theta = 60.0^o$

The depth of the pool is $d = 3.00 \ m$

Generally the velocity of the stone towards the horizontal direction is

$v_h = v * cos (\theta)$

=> v_h = 4.00 * cos (60.0)

=> $v_h = 2 \ m/s$

Generally the velocity of the stone towards the vertical direction is

$v_v = v * sin (\theta)$

=> v_v = 4.00 * sin (60.0)

=> $v_v = 3.46 \ m/s$

Generally from kinematic equation we have that

$s = v_v t + \frac{1}{2} gt^2$

Here s = -h , the reason for the negative sign is because the distance traveled is towards the negative y-axis

t is the time taken by the stone to travel through air

$-2.50= 3.46 t + \frac{1}{2} (-9.8)t^2$

Here g = -9.8 m/s^2 because the direction of movement is in the negative y axis

=> 4.9 t^2 -3.46t - 2.50

Solving the above equation using quadratic formula we obtain t to be

t = 1.15 s

Generally from kinematic equation

v_j = v_v +gt

Here

is the velocity of the stone just before it hits the water

=> v_j = 3.46 + -9.8(1.15)

=> v_j = -7.81 m/s

The negative sign just signify's the direction of the velocity

The magnitude is

|v_j| = 7.81 m/s

Generally the speed of the stone in water is

$v_w = \frac{|v_j|}{2}$

=> $v_w = \frac{7.81 }{2}$

=> $v_w = 3.905 \ m/s$

The time taken to travel through the water is

$t_w = \frac{d}{v_w}$

=> $t_w = \frac{3}{3.905}$

=> $t_w = 0.7682 \ s$

Generally the total time taken is mathematically represented as

t_t = t + t_w

=> t_t = 1.15 + 0.7682

=> $t_t = 1.9182 \ s$

Answer from: Quest

newton's first law of motion

Answer from: Quest

increase masses, decrease separation

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