Question

baseball is seen to pass upward by a window 25 m above the street with a vertical speed of 15 m/s. The ball was thrown from the street.
(a) What was its initial speed?
1 m/s
(b) What altitude does it reach?
2 m
(c) How long after it was thrown did it pass the window?
3 s
(d) After how many more seconds does it reach the street again?
4 s

Answer 1

(a)from the equation of motion v^2-u^2=2aShere the final velocity v =15 m/s , height of the window is S =25 macceleration a = -g here g is acceleration due to gravity , -ve sign represents due to upward motionu^2 =v^2+2gS=15^2+2*9.8*25initial velocity of the ball at ground is u =26.73 m/s_____________________________________________________________________________(b)the equation for maximum height is H =u^2/2g=(26.73)^2/2*9.8=36.47m_____________________________________________________________________________(c)from the equation of motionv = u - gttime taken t =(u-v)/g=(26.73 -15)/9.8=1.196 s_____________________________________________________________________________(d)let the extra time taken is =t₁from the equation of motion v=u+atherefinal velocity is - v and total time = t+t₁-v =u-g(t+t₁)g(t+t₁)=u+vt+t₁=(u+v)/g=(15+26.73)/9.8=4.2581 st₁=4.2581 s-t=4.2581 s- 1.196 s=3.062 s