Question

You are playing right field for the baseball team. Your team is up by one run in the botton of the last inning of the game when a ground ball slips through the infield and comes straight toward you. As you pick up the ball 65m from home plate, you see a runner rounding third base and heading for home with the tying run. ou throw the ball at an angle of 30 degrees above the horizontal with just the right speed so that the ball is caught by the catcher, standing on home plate, at the same height as you threw it. As you release the ball, the runner is 20 meters from home plate and running full speed at 8.0 m/s. Will the ball arrive in time for your team's catcher?

Answer 1

Answer:

The correct response is "No". The further explanation is given below.

Explanation:

The given values are:

Angle

$\theta = 30^{\circ}$

Distance

d = 20 m

Speed

s = 8.0 m/s

Now,

⇒  $y(t) = \frac{1}{2}\times a\times t^2 + v0\times t\times (sin \theta)$

    $0 = -4.9t^2 + v0\times t\times (sin 30)$

    $x = v0\times (cos \theta)\times t$

    $65 = v0\times (cos 30)\times t$

    $v0\times t = \frac{65}{Cos30}$

    $0 = -4.9t^2 + 65\times \frac{sin30}{cos30}$

    $t = 2.767 \ sec$

So,

⇒  $d = r\times t$

    $20 = 8\times t$

      $t=2.5 \ sec$

Therefore, 2.5 < 2.767, so it won't get there.