Question

A ball thrown vertically upwards with a velocity of 15 m/s reaches its highest point at 11.25 m in 1.5 sec. Find the total distance travelled by the ball and its position after 2 sec respectively. Take $g=10m/s^2$

• A. Total distance = 12 m, After 2 sec = 9.5 m.
• B. Total distance = 17.5 m, After 2 sec =7.5 m.
• C. Total distance = 14 m, After 2 sec = 9 m.
• D. Total distance = 12.50 m, After 2 sec = 10 m.

Answer 1

The correct option is D Total distance = 12.50 m, After 2 sec = 10 m.

The ball is thrown upwards.
$\therefore$ Acceleration due to gravity $=-10m/s^2$
Final velocity $=0m/s$
Initial velocity $=15m/s$
Distance $=11.25m$
Time $=1.5sec$
To find total distance traveled:

the distance traveled by the ball in upward and downward motion will be the sum of the height reached upto 1.5 sec plus height descended in next 0.5 sec.

Height descended,

$h=ut+\frac{1}{2}a{t}^2$

$h=0+\frac{1}{2}(-10)(0.5{)}^2$

$h=1.25m$

Hence, the total distance will be $11.25m+1.25m=12.50m$

To find position after $2sec$
Using the formula:
$s=ut+\frac{1}{2}a{t}^2$
$s=15\times 2+\frac{1}{2}\times -10\times (2{)}^2$
$s=30+(-40)$
$s=-10m$

So the option D is correct.