Question

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?

Answer 1

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?

Answer:

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

 So

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

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 2

The total time taken by the stone to reach the bottom of the pool is 1.92s

The motion of the stone to the bottom of the pool:

Given information:

height above the water surface from which the stone is thrown, h = 2.5m

the initial velocity of the stone, u = 4 m/s

the angle from the horizontal θ = 60°

The vertical velocity of the stone in given by:

u' = usin60

u' = 4sin60

u' = 3.46 m/s

(i) The time taken by the stone to reach the water surface:

From the second equation of motion we get:

h = ut + ¹/₂gt²

the stone is thrown up at an angle of 60°, so the vertical velocity is upwards, the final position of the stone is at the surface of the water, which is 2.5m downwards from the position it is thrown, also the acceleration due to gravity is downwards. So if we consider upwards as a positive direction and downwards as a negative direction, we get:

-2.5 = 3.46t - 0.5×9.8t²

-2.5 = 3.46t - 4.9t²

solving the above we get:

t = 1.15 s

The vertical speed of stone when it touches the surface of water;

v = u + gt

v = 3.46 - 9.8×1.15

v = -7.81 m/s

or, 7.81 m/s downwards

(ii) Motion of the stone to the bottom of the pool:

In this case, the motion is in a straight line, and the stone descends with constant speed which is half the speed of the stone when it touches the surface of the water, so:

u = -7.81/2

u = -3.9 m/s

The depth of the pool is d = -3m

so time taken:

t' = d/u = (-3)/(-3.9)

t' = 0.77 s

So the total time taken is:

T = t + t'

T = 1.15 + 0.77

T = 1.92 s

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