thrown at an angle of 30.0° with the horizontal. At what speed must the second b

Question

thrown at an angle of 30.0° with the horizontal. At what speed must the second ball be thrown so that it reaches the same height as the one thrown vertically? Problem 2 (Newton's Second Law A boat moves through the water with two forces acting on it. One is a 2 000-N forward push by the water on the pro- peller, water around the bow. (a) What is the acceleration of the 1000-kg boat? (b) If it starts from rest, how far will the boat move in 10.0 s? (c) What will its velocity be at the end of that time? and the other is a 1 800-N resistive force due to the Problem 3 (Statics) (a) Find the tension in each cable supporting the 37.0 600-N cat burglar (b) Suppose the hori- zontal cable were reattached

Explanation / Answer

1.

In case1 when ball is thrown vertically upward

given time taken = 3.0 sec

time to reach max height = total time/2 = 3/2 = 1.5 sec

Suppose initial velocity is = Vi

then using equation

Vf = Vi + a*t

Vf = speed at max height = 0 m/sec

0 = Vi - 9.8*1.5

Vi = 9.8*1.5 = 14.7 m/sec

Now max height will be

Vf^2 = Vi^2 + 2*a*h

h = (0^2 - 14.7^2)/(2*(-9.8)) = 11.03 m

Now in case 2:

In projectile motion max height is given by:

Hmax = V0^2*(sin theta)^2/2g

V0 = sqrt (Hmax*2*g/(sin theta)^2)

V0 = sqrt (11.03*2*9.81/(sin 30 deg)^2)

V0 = 29.42 m/sec

So second ball must be thrown at 29.42 m/sec

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