Rex Things throws his mother’s crystal vase vertically upwards with an initial velocity of \[26.2m/s\]. Determine the height to which the vase will rise above its initial height.
A) $34.0m$
B) $39.0m$
C) $35.0m$
D) $37.0m$

Hint: The vase is thrown vertically upwards hence its height will increase only in the vertical direction. Recall the formula to calculate maximum height during a projectile motion. Also the angle here is \[{{90}^{\circ }}\].


Formula Used:
The formula to calculate maximum height during a projectile motion is given by
\[{{H}_{\max }}=\dfrac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\]
Where
\[{{H}_{\max }}\] Represents maximum height
\[u\] Represents initial velocity
\[\theta \] Represents the angle at which the object is thrown with respect to horizontal
\[g\]Represents the acceleration due to gravity

Complete step by step answer:
Since the object is thrown vertically, the angle \[\theta \] will be \[{{90}^{\circ }}\]
To determine the height to which the vase will rise above its initial height, we will use the above mentioned formula
\[{{H}_{\max }}=\dfrac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\]
The given values are initial velocity (\[u\] ) \[=\] \[26.2m/s\], \[g=9.8m/{{s}^{2}}\], \[\theta ={{90}^{\circ }}\]
Putting these values in the above formula, we get
$\Rightarrow {{H}_{\max }}=\dfrac{{{(26.2)}^{2}}({{\sin }^{2}}{{90}^{\circ }})}{2(9.8)}$
\[\Rightarrow {{H}_{\max }}=\dfrac{(686.44)(1)}{19.6}\]
\[\Rightarrow {{H}_{\max }}=35.02m\]
When we round off this value to single decimal then the answer is \[35.0m\] which is option \[C\]

Note: In the given question the angle was not given but it is known that during vertically upwards motion the angle with the horizontal is \[{{90}^{\circ }}\].
Also remember that for calculating range we have a different formula,
\[R=\dfrac{{{u}^{2}}\sin 2\theta }{g}\]
Where \[R\]is the range
\[u\] Is the initial velocity
\[\theta \] Is the angle of projection with the horizontal
\[g\] Is the acceleration due to gravity
Also to increase the range in vertically upward motion we have to increase the initial velocity to get maximum height. And Range is maximum when the angle of projection \[\theta ={{90}^{\circ }}\]
\[\theta ={{45}^{\circ }}\]. Do remember that Range is always concerned with horizontal direction and Height is always concerned with vertical direction.