Question

A rock is thrown from a height of 2.0 m at a window that is located 9.0 m above the ground. The initial velocity of the rock is 20 m/s, and it is thrown at an angle of 40 above the horizontal. If the rock strikes the window on the upward trajectory, what is the horizontal distance between the person throwing the rock and the window

Answer 1

Answer:

The value is  $x = 11.81 \ m$

Explanation:

From the question we are told that

   The height is h =  2.0 m

   The height of the window is $d = 9.0 \ m$

    The initial velocity of the rock is $u = 20 \ m/s$

    The angle at which it is thrown is $\theta = 40$

Generally the  vertical component of the velocity of the stone is mathematically represented as

    $v_y = 20 sin (40)$

=>$v_y = 12.86 \ m/s$

Generally the height of the window from the ground is mathematically represented as using kinematic equation as

      $d = h + v_yt + \frac{1}{2} gt^2$

=>   $9 = 2 +12.86 t + \frac{1}{2} * - 9.8 t^2$

Here g is negative -9.8 m/s^2 because the direction of the stone is against gravity

   So

       $4.9 t^2 -12.86 t + 7 =0$

Solving this quadratic equation using quadratic formula we obtain

     t = 0.770 s

Generally the velocity of the stone on the x axis is mathematically represented as

       $v_x = 20 * cos(40 )$

=>    $v_x = 15.32 \ m/s$

Generally the distance between the person throwing the rock and the window is mathematically represented as

       $x = v_x * t$

=>    $x = 15.32 * 0.771$

=>    $x = 11.81 \ m$