A)A ball is thrown straight upward and returns to the thrower\'shand after 3.40

Question

A)A ball is thrown straight upward and returns to the thrower'shand after 3.40 s in the air. A secondball is thrown at an angle of 33.0° with thehorizontal. At what speed must the second ball be thrown so that itreaches the same height as the one thrown vertically?
1 m/s


B)How long does it take an automobile traveling in the left lane ofa highway at 70.0 km/h to overtake (becomeeven with) another car that is traveling in the right lane at35.0 km/h when the cars' front bumpers areinitially 115 m apart?
1 s

C)A brick is thrown upward from the top of a building at an angleof 20° to thehorizontal and with an initial speed of 11m/s. If the brick is in flight for 2.8 s,how tall is the building?
1 m

D)A tennis player standing 12.4 m from thenet hits the ball at 3.40° above the horizontal. To clear the net, the ballmust rise at least 0.332 m. If the balljust clears the net at the apex of its trajectory, how fast was theball moving when it left the racquet?
1 m/s

Explanation / Answer

(A). Time of flight T = 3.4 s                     2v / g = 3.4 s from this speed v = 3.4 g / 2                            = 16.66 m / s Maximum height reached H = v ^ 2 / 2g                                           = 14.161 m Angle of second ball = 33.0° maximum height of 2 nd ball H ' = H = 14.161 m we know H ' = ( v sin ) ^ 2 / 2g from this v sin = [ 2gH ]                         = 16.66 from this speed v = 16.66 / sin 33                           = 30.58 m / s (B). speed of car in left lane v = 70.0 km/h speed of the car in the right lane v ' = 35.0 km/h Separation S = 115 m relative speed   V = v - v '= 35 km / h                           = 35 * ( 5/ 18 ) m / s                           = 9.7222m / s So, required time t = S / V                             = 11.828 s

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