A ball is thrown vertically upwards with a velocity of 98 metre per second calculate a maximum height it rises and the total time it takes to reach the ground

Dear Student

Initial Velocity, u = 98 m/s
Acceleration due to gravity, g = - 9.8 m/s²

Now, maximum height it rises can be calculated as -
At maximum height, velocity = 0
Therefore, using -
v² = u² + 2as
0 = (98)² + 2 (-9.8)s
0 = 9604 - 19.6 s
19.6 s = 9604

So, s = $\frac{9604}{19.6}$ = 490 m
490 m is the maximum height reached.

Now, the total time taken to reach the ground can be calculated as -
Since the net displacement becomes zero
So, using -
s = ut + $\frac{1}{2}$ at²         $\Rightarrow$   0 = 98 t + ​$\frac{1}{2}$ (-9.8) t² 
0 = 98 t - ​$\frac{1}{2}$ x 9.8 t²      $\Rightarrow$  ​4.9 t² = 98 t    $\Rightarrow$ 4.9 t = 98
t = $\frac{98}{4.9}$ = 20 s
20 sec is the time taken to reach the ground.


Regards