Question

a ball is thrown vertically upward with a velocity of 48 m/s calculate 1. the maximum height to which it rises . 2. the total time it takes to return to the surface of the earth
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Answer 1

Explanation:

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Answer 2

Answer:

maximum height to which ball rises is 117.5 m.time taken to reach the surface of earth is 4.8 sec.

Explanation:

we will be using equation of motion

Given : ball thrown with initial speed (u) 48m/s

Step 1 : When ball rises to maximum height final velocity of ball will be 0               i.e.  v = 0

 using

 $v^{2} - u^{2} = 2as$        

      $0 - 48^{2} = 2(-9.8)s$      

         $\frac{2304}{19.6} = s$    

    $s = 117.5 m$        

    hence maximum height obtained is 117.5 m

Step 2 : total time taken to return to surface of earth can be calculated using  

   $v= u + at$  

 where u = 0

and  

$v^{2} -0 = 2 (9.8)(-117.5)$

 $v^{2} = 2303$  

 $v = 47.98$

now ,

 $v = u + at$

 $47.98 = (9.8)t$  

$t= \frac{47.98}{9.8}$  

$t = 4.89 sec$