A cricket ball is thrown at a speed of 30m/s in a direction of 30 degrees above the horizontal. calculate:
a)the maximum height
b)the distance from the thrower to where the ball returns to the same level
c)the time taken to return to the same level
appreciate your help :)
a)the maximum height
b)the distance from the thrower to where the ball returns to the same level
c)the time taken to return to the same level
appreciate your help :)
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a) Hmax = Uy^2/2g = (30*sin(radians(30)))^2/(2*9.81) = 11.46788991 = 11.5 m above release height. ANS.
b) y(T) = 0 = Uy T - 1/2 gT^2; so T = sqrt((2Uy/g) = ? seconds time to return to release height. Then X(T) = Ux T = U cos(30) sqrt(2Uy/g) = 30*cos(radians(30))*sqrt(2*30*sin(radian… = 45.43368996 = 45.4 m from release point. ANS.
c) T = sqrt(2*30*sin(radians(30))/9.81) = 1.748743542 = 1.75 seconds from time of release. ANS.
b) y(T) = 0 = Uy T - 1/2 gT^2; so T = sqrt((2Uy/g) = ? seconds time to return to release height. Then X(T) = Ux T = U cos(30) sqrt(2Uy/g) = 30*cos(radians(30))*sqrt(2*30*sin(radian… = 45.43368996 = 45.4 m from release point. ANS.
c) T = sqrt(2*30*sin(radians(30))/9.81) = 1.748743542 = 1.75 seconds from time of release. ANS.
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Do your own homework and stop using the internet as your crutch...Think your clever do you? The answers in your book on page 48...
(haha just kidding)
(haha just kidding)