Question

A ball was thrown from height H and the ball hit the floor with velocity `10(hati-hatj)` after 1.5 sec of its projection. find initial velocity of ball?

Answer 1

Given that ball hit the floor with velocity `10hati-10hatj`m/s
The horizontal component of the velocity of projection remains unaltered during the flight of the projectile.

If we consider that the vertical component of velocity of projection is `uhatj` then the velocity of projection will be `10hati+uhatj`m/s

Applying equation of kinematics for the change in vertical component of velocity during time of flight `1.5s` we get

`-10=u-gxx1.5`

`=>-10=u-10xx1.5`

`=>u=5`m/s

So initial velocity of the ball `10hati+uhatj` m/s

`=10hati+5hatj` m/s

Its magnitude `sqrt(10^2+5^2)=5sqrt5` m/s

Direction of projection `tan^-1(5/10)=tan^-1(1/2)=26.5^@` with the horizontal

Answer 2

`v=(v_(x_0),v_(y_0)) = (10,4.715)`

Explanation:

The parametric movement equations are:

`{(x = x_0+v_(x_0)t),(y=y_0+v_(y_0)t-1/2 g t^2):}`

and the velocities

`{(dot x=v_(x_0)),(dot y = v_(y_0)-g t):}`

After `Delta t` seconds we have

`{(dot x_(Delta t) = v_(x_0)),(dot y_(Delta t) = v_(y_0)-g Delta t):}`

or

`{(10=v_(x_0)),(-10=v_(y_0)-9.81*1.5):}`

then the initial velocity is

`v=(v_(x_0),v_(y_0)) = (10,4.715)`