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?

2 Answers
Jul 8, 2017

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

Jul 8, 2017

#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)#