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

2 Answers
Jul 8, 2017

Given that ball hit the floor with velocity 10hati-10hatj10ˆi10ˆjm/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 uhatjuˆj then the velocity of projection will be 10hati+uhatj10ˆi+uˆjm/s

Applying equation of kinematics for the change in vertical component of velocity during time of flight 1.5s1.5s we get

-10=u-gxx1.510=ug×1.5

=>-10=u-10xx1.510=u10×1.5

=>u=5u=5m/s

So initial velocity of the ball 10hati+uhatj10ˆi+uˆj m/s

=10hati+5hatj=10ˆi+5ˆj m/s

Its magnitude sqrt(10^2+5^2)=5sqrt5 102+52=55 m/s

Direction of projection tan^-1(5/10)=tan^-1(1/2)=26.5^@tan1(510)=tan1(12)=26.5 with the horizontal

Jul 8, 2017

v=(v_(x_0),v_(y_0)) = (10,4.715)v=(vx0,vy0)=(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)