Kinematics - Finding velocity of a baseball

AI Thread Summary
A baseball player hits a ball that lands 24 meters above the field at a velocity of 50 m/s at a 35-degree angle below the horizontal. To find the initial velocity when hit from 1 meter above the field, calculations show it to be approximately 50.61 m/s. The horizontal distance traveled by the ball is calculated to be around 235 meters. The time the ball is in the air is determined to be approximately 5.74 seconds. The discussion emphasizes using kinematic equations to solve for velocity, distance, and time in projectile motion scenarios.
kidia
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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?
 
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Let the horizontal and vertical components of the initial velocity be u_1, u_2

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

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

So inital velocity = \sqrt {(u_1^2 + u_2 ^2)}

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

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