a baseball is thrown at an angle of 53.1 degrees above the horizontal with an initial velocity of 50.0 m/s

a) at what two times is the baseball at a height of 25.0m above the point from which it was thrown?

b) calculate the horizontal and verticle components of the baseballs velocity at each of the two times calculated in part a

c) what are the magnitude and direction of the baseballs velocity when it returns to the level from which it was thrown?

a) at what two times is the baseball at a height of 25.0m above the point from which it was thrown?

b) calculate the horizontal and verticle components of the baseballs velocity at each of the two times calculated in part a

c) what are the magnitude and direction of the baseballs velocity when it returns to the level from which it was thrown?

Guest Sep 26, 2013

#1**0 **

Initially y dash = 50*sin 53.1 degrees x dash = 50*cos 53.1 degrees

y=0 x=0

ongoing

y"=-g (I'll use -10) x"=0 ignore wind resistance

y' = -10t + 50sin 53.1 x'= 50*cos53.1

y= -5t^2 +(50sin53.1)*t x= (50cos53.1) *t

When y = 25

just substitute into the appropriate equation and solve with a quadratic equation.

I got 7.31 sec and 0.68 secs but I might have made a mistake.

This is enough to get you started. hope it helps.

y=0 x=0

ongoing

y"=-g (I'll use -10) x"=0 ignore wind resistance

y' = -10t + 50sin 53.1 x'= 50*cos53.1

y= -5t^2 +(50sin53.1)*t x= (50cos53.1) *t

When y = 25

just substitute into the appropriate equation and solve with a quadratic equation.

I got 7.31 sec and 0.68 secs but I might have made a mistake.

This is enough to get you started. hope it helps.

Melody Sep 28, 2013