Kinematics - Finding velocity of a baseball

  • Thread starter kidia
  • Start date
  • #1
66
0
Hi I have a question
A baseball player hits a baseball that lands in stands 24 meters above the playing field.The ball land with velocity of 50 meters per second at angle of 35 degree below the horizontal.

a.If a player hits the ball 1 meter above the playing field,what was the velocity of the ball upon leaving the bat?
b.What was the horizontal distance traveled by ball
c.how long was the ball in air

I have done like this:

let r be angle in degrees

t=2usinr/g

time=?
velocity=50m/s
gravity=10

time= 2*50*sin35/10=5.74s
velocty of ball at 1m will be:

y=vsinrt-0.5gt
1=usin35*5.74-0.5*10*5.74*5.74
u=50.61m/s


b.horizontal distance traveled by ball

distance/time=velocity
then
distance/time=ucosr
distance=50cos35*5.74
distance=235m

c.Time of the ball in air.

t=2usinr/g

time=?
unitial velocity=50m/s
gravity=10

time= 2*50*sin35/10=5.74s

Any idea about this question?
 

Answers and Replies

  • #2
357
11
Let the horizontal and vertical components of the initial velocity be [tex]u_1, u_2[/tex]

Apply [tex] v^2 = u^2 + 2fS[/tex] in vertical and horizonatal directions to find u1 and u2.

Note that the vertical distance travelled is 24 -1 = 23 m.

So inital velocity = [tex]\sqrt {(u_1^2 + u_2 ^2)}[/tex]

(2). Apply [tex] v = u + ft [/tex] vertically to find the time of travell

(3) Apply [tex] s = ut + \frac{1}{2} f t^2 [/tex] horizontally to find the distance travelled.
 
  • #3
66
0
Ooo is fine but my doubt is about the angle 35 degees u didn't mention it
 

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