A baseball is bunted down the thired-base line leaving the bat from a height of

Question

A baseball is bunted down the thired-base line leaving the bat from a height of 0.900m with an initial velocity of 3.00m/s at an angle of 20.0° above the horizontal.

A) How far does the baseball travel before hitting the ground?
B) what is the speed of the ball just before it hits the ground?
C) what is the maximum height of the baseball above the horizontal in this travel?
D) what is the ball normal (centripetal) acceleration at the maximum height ?
E) what is the baseball tangential acceleration at the maximum height ?

Explanation / Answer

Given: initial velocity of the baseball u = 3.00 m/s angel = 200 ........................................................................ time taken to reach the ball to ground -h = usin - 1/2 gt2 ..... (1) here the negative sign indicates the ball moves downwards. t = sqrt [2*(usin+h)]/g]    = sqrt [2*(3.00m/s)sin200+0.9 m)]/9.8 m/sec2]    = 0.45 sec ............................................................................................ now (a) Range R = ux*t                = (3.00)cos200*0.45 sec                = 1.26 m (b) final velocity before reaches the ground vertical velocity at given point vy = sqrt[(usin)2 +2gh]                                                        = sqrt[(3.00m/s*sin(200))2 +2*9.8m/sec2 (0.900 m)]                                                        = 4.32 m/s so final velocity when touches the ground v = sqrt[(ucos)2 +(vy)2]                                                                       = sqrt[(14.12)+(18.66)]                                                                       = 5.72 m/s (c) maximum height reached H = u2sin2/2g                                             = (3.00 m/s)2(sin 200) 2/ 2*9.8 m/sec2                                             = 0.05 m (d) acceleration a = v2/r                         = u2cos2 /r                          = (3.00 m )2(cos200)2 / 0.05 m                                               = 0.05 m (d) acceleration a = v2/r                         = u2cos2 /r                          = (3.00 m )2(cos200)2 / 0.05 m                            = 158 m /sec2 (e) Tangential acceleration at the maximum height of the projectile is zero. (e) Tangential acceleration at the maximum height of the projectile is zero.

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