Question

A baseball is thrown down ward from 15m tower with an initial speed of 5m/s . Determine the speed at Which it hits the ground and the time of travel

Answer 1

Answer:

Hope it helps

Answer 2

The speed at which the baseball hits the ground is 5√13 i.e., 18.02 m/s and the time of travel of the baseball is  $\frac{\sqrt{13} -1}{2}$ i.e., 1.302 seconds.

Given:

The height from which the baseball is thrown downwards is = (s) = 15 m

The initial downward speed of the baseball is (u) = 5 m/s.

To Find:

1. The speed with which it strikes the ground (v) =?

2. The time of travel of the baseball (t) =?

Solution:

Here, we will use the 2nd kinematical equation to find the time of flight and the 1st kinematical equation to calculate its final velocity of striking.

Let us assume all the quantities measured in the downward direction as negative and all quantities measured in the upward direction as positive.

Hence, s = -15 and u = -5 ;

The acceleration due to gravity (a) = g = -10  

Now, The second kinematical equation is as follows;

i.e.,  s = $ut + \frac{1}{2}at^{2}$

By Substituting all the values we get;

⇒  -15 = -5(t) + (0.5)(-10)$t^{2}$

By dividing the equation by -5 we get;

⇒  3 = t + $t^{2}$

⇒   $t^{2}$ + t - 3 = 0

∴ t =  $\frac{-1+\sqrt{13} }{2}$  or  t = $\frac{-1-\sqrt{13} }{2}$

∵ time (t) cannot be negative, ∴ t ≠  $\frac{-1-\sqrt{13} }{2}$

∴ t =  $\frac{-1+\sqrt{13} }{2}$  seconds.

Now, using the first kinematical equation as follows;

i.e.,  v = u + at

⇒    v = -5 + (-10)( $\frac{-1+\sqrt{13} }{2}$ )

⇒    v = -5 + (-5)($-1 + \sqrt{13}$ )

⇒    v = -5 +5 -5√13

⇒    v = -5√13 m/s

Here, the negative sign denotes the direction of the final velocity, i.e., it is directed downwards.

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