A batter hits a baseball so that it leaves that bat with the initial velocity 37 m/s at an angle of 53° with horizontal. Find the position of the ball, magnitude and direction of the velocity of the ball after 2 sec. Treat the base ball as a projectile.
Answers
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
=
2
s
So,
h
=
39.5
m
And horizontal displacement will be
r
=
37
cos
53
⋅
2
=
44.52
m
So,after
2
s
the baseball will be lying
39.5
m
above its point of projection and
44.52
m
ahead of its point of projection.
Now,let the vertical component of velocity will become
v
y
after time
2
s
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
=
22.27
m
s
So,magnitude of velocity after
2
s
is
√
v
2
x
+
v
2
y
=
24.4
m
s
Making an angle of
tan
−
1
(
9.95
22.27
)
=
24.06
∘
w.r.t horizontal