Question

A place-kicker must kick a football from a point 36.0 m (about 40 yards) from the goal. Half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 21.6 m/s at an angle of 50.0° to the horizontal.

Required:
By how much does the ball clear or fall short (vertically) of clearing the crossbar?

Answer 1

Answer:

The difference is height is $\Delta h =6.92 \ m$

Explanation:

From the question we are told that

     The distance of ball  from the goal is $d = 36.0 \ m$

    The height of the crossbar is  $h = 3.05 \ m$

       The speed of the ball is $v = 21.6 \ m/s$

       The angle at which the ball was kicked is $\theta = 50 ^o$

The height attained by the ball is mathematically represented as

      $H = v_v * t - \frac{1}{2} gt^2$

Where $v_v$ is the vertical component of  velocity which is mathematically represented as

     $v_v = v * sin (\theta )$

substituting values

     $v_v = 21.6 * (sin (50 ))$

     $v_v = 16.55 \ m/s$

Now the time taken is  evaluated as

       $t = \frac{d}{v * cos(\theta )}$

substituting value

     $t = \frac{36}{21.6 * cos(50 )}$

    $t = 2.593 \ s$

So

     $H = 16.55 * 2.593 - \frac{1}{2} * 9.8 * (2.593)^3$

     $H = 9.97 \ m$

The difference  in height is mathematically evaluated as

      $\Delta h = H - h$

substituting value

    $\Delta h = 9.97 - 3.05$

    $\Delta h =6.92 \ m$