"Person throws a rock vertically up to the sky from 20 meter tower. The starting speed of the rock is 30m/s. Air resistance is neglected.?

a) Calculate the maximum altitude of the rock and the time that it takes to get the max altitude before starting to fall down. 

b)Calculate the time starting from throwing the rock and ending with reaching the ground.(The whole time of throwing the rock, including falling down). Calculate the final speed of the rock when its going to hit the ground.

c)Calculate the rocks total summary mechanical energy while throwing the rock, at max altitude, at the moment when it hits the ground."

1 Answer

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  • 5 months ago

    a) let the rock reaches the maximum height in t sec

    =>BY v = u - gt

    =>0 = 30 - 9.8 x t

    =>t = 30/9.8 = 3.06 sec

    Let the height attained in t sec is h meter

    =>By s = ut - 1/2gt^2

    =>h = 30 x 3.06 - 1/2 x 9.8 x (3.06)^2

    =>h = 45.88 m

    Thus the total height from the ground (H) = h + 20 = 65.88 m

    (b) let the rock take t sec to fall H meter,

    =>By s = ut + 1/2gt^2

    =>65.88 = 0 + 1/2 x 9.8 x t^2

    =>t = √13.44 = 3.67 sec

    Thus the total time of the flight (T) = t + 3.06 = 6.73 sec

    Let the velocity after a H meter fall is v m/s

    Thus by v^2 = u^2 + 2gH

    =>v = 0 + 2 x 9.8 x 65.88

    =>v = √1291.25 = 35.93 m/s

    (c) By the law of energy conservation:

    =>PE(initial) + KE(initial) = PE(Top) = KE(final) = 1/2 x m x (35.93)^2 = 645.62m Jule

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