Question

A batter hits a baseball so that it leaves the bat with an initial speed 37 m/s at an angle of 53.what will be the position of the ball and the magnitude and direction of its velocity after 2 seconds?Treat the baseball as a projectile.

Answer 1

As,we know,in case of a projectile motion,the projectile moves forward with its constant horizontal component of velocity and after going for a certain height due to its vertical component of velocity it reaches back to the ground due to gravity.

So,we can say after `t` if it reaches height `h` then,

`h=(37 sin 53)t - 1/2 *9.8 t^2` (as,vertical component of velocity is `37 sin 53`)

Given, `t=2s`

So, `h=39.5m`

And horizontal displacement will be `r=37 cos 53*2=44.52m`

So,after `2s` the baseball will be lying `39.5m` above its point of projection and `44.52m` ahead of its point of projection.

Now,let the vertical component of velocity will become `v_y` after time `2s`

So, `v_y=37 sin 53 -9.8*2`

or, `v_y=9.95 m/s`

And,horizontal component of velocity remains constant i.e `v_x=37 cos 53=22.27m/s`

So,magnitude of velocity after `2s` is `sqrt(v_x ^2 + v_y ^2)=24.4m/s`

Making an angle of `tan^-1(9.95/22.27)=24.06^@` w.r.t horizontal