Question

Positions B and D are 11.3 meters and 1.9 meters above the playing field, respectively. If the ball had a speed of 6.2 m/s in position B, what is its speed in position D?

Answer 1

Answer:

Speed of the ball in position D $=14.92 \frac{m}{s}$

Explanation:

Given:

Position of B=11.3$\text { meters }$

Position of D=1.9$\text { meters }$

Velocity of B=6.2$\text { meters }$

To Find:

Velocity of D

Solution:

According to the formula, Velocity is given as

$V d=\sqrt{\left[V b^{2}+(2 \times g \times d y)\right]}$

$V b$=Velocity of B

$V d$=Velocity of D

g=acceleration due to gravity=9.8 m/s^2  

$d y$=Change in position of B and D

Substitute the all values in the above equation we get  

$d y=11.3-1.9$

$V d=\sqrt{\left[6.2^{2}+(2 \times 9.8 \times(11.3-1.9))\right]}$

$=\sqrt{[38.44+(2 \times 9.8 \times(9.4))]}$

$=\sqrt{[38.44+(19.6 \times 9.4)]}$

$=\sqrt{38.44+186.24}$

$=\sqrt{222.68}$

$=14.92 \frac{m}{s}$

Result :

The velocity of D is $=14.92 \frac{m}{s}$