Q.1 a. How far from the dog will the suitcase land? You canignore air resistance
ID: 1737023 • Letter: Q
Question
Q.1 a. How far from the dog will the suitcase land? You canignore air resistance. Give your answers in ascending orderseparated with comma. b. Calculate the horizontal &vertical component of the baseball's velocity at anearlier time calculated in part(a). c.Calculate the horizontal &vertical component of the baseball's velocity at a later time calculated in part (a). d. What is the magnitude of the baseball'svelocity when it returns to the level at which it left thebat? e. What is the direction of the baseball'svelocity when it returns to the level at which it left thebat? d. What is the magnitude of the baseball'svelocity when it returns to the level at which it left thebat? e. What is the direction of the baseball'svelocity when it returns to the level at which it left thebat?Explanation / Answer
Q.1The velocity of the airplane is u = 88.0 m/s The airplane is flying at an angle of 21.0oabove the horizontal The range of the airplane is R = (u2 * sin2/g) --------------(1) The maximum height of the airplane is H = (u2 * sin2/2g) or sin2 = (2gH/u2) or sin = ((2gH)1/2/u) or = sin-1((2gH)1/2/u)----------------(2) Here,g = 9.8 m/s2 and H = 116 m or = sin-1((2 * 9.8 *116)1/2/(88.0)) or = 32.81o Substituting the above value of in equation (1),weget R = ((88.0)2 * sin(2 *32.81o)/9.8) or R = 719.74 m The distance from the dog the suitcase will land is R = 719.74m. Q.2The speed of the baseball when it leaves the bat is u =29.7 m/s The baseball leaves the bat at an angle of =38.0o above the horizontal The maximum height reached by the baseball is H = 10.1 m a.From equation (2),we get = sin-1((2gH)1/2/u) or = sin-1 ((2 * 9.8 *10.1)1/2/(29.7)) or = 28.27o The maximum distance travelled by the baseball is R = (u2 * sin2/g) or R = ((29.7)2 * sin(2 *28.27o)/(9.8)) or R = 75.1 m We know from the relation R = ut + (1/2)gt2 or 75.1 = 29.7t + (1/2) * 9.8 * t2 or 4.9t2 + 29.7t - 75.1 = 0 Solving the above quadratic equation,we get t1= 1.92 s and t2= -7.98 s Therefore,the two times at which the baseball reaches a heightof 10.1 m above the point at which it left the bat are t1= 1.92 s and t2= -7.98 s. b.The velocity of the baseball at an earlier time calculatedin part (a) is v = u + gt1 or v = 29.7 + 9.8 * 1.92 = 48.52 Therefore,the horizontal & vertical component of thebaseball's velocity at an earlier time calculated in part (a)is vx= v * cos = 48.52 * cos(28.27o)= 42.73 m/s and vy= v * sin = 48.52 *sin(28.27o) = 22.98 m/s c.The velocity of the baseball at a later time calculatedin part (a) is v = u + gt2 or v = 29.7 + 9.8 * (-7.98) = 48.504 m/s Therefore,the horizontal & vertical component of thebaseball's velocity at an earlier time calculated in part (a)is vx= v * cos = 48.504 * cos(28.27o)= 42.72 m/s and vy= v * sin = 48.504 *sin(28.27o) = 22.97 m/s d.Let the magnitude of the baseball's velocity when itreturns to the level at which it left the bat be v.Therefore,weget v2 - u2 = 2gR or v2 = u2 + 2gR or v = (u2 + 2gR)1/2 or v = ((29.7)2 + 2 * 9.8 *75.1)1/2 or v = 48.52 m/s e.The direction of the baseball's velocity when it returns tothe level at which it left the bat is t1= (2u * sin/g) or sin = (t1 * g/2u) or = sin-1(t1 * g/2u) or = sin-1(1.92 * 9.8/2 * 29.7) or = 18.46oRelated Questions
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