A stone is thrown from a cliff top with a vertical velocity of 50 m/s and horizo

Question

A stone is thrown from a cliff top with a vertical velocity of 50 m/s and horizontal velocity of 20 m/s.

A) What is its resultant velocity?

B) At what angle above the horizontal is it thrown?

C) What is the horizontal velocity at its highest point? What is its vertical velocity? Show a calcualtion or explain your answer.

D) What is its horizontal velocity after 8 seconds? What is its vertical velocity? Its resultant (vector sum of x and y components) velocity?

E) How far has the stone traveled horizontally in that time?

Explanation / Answer

(a) resultant velocity = sqrt(50^2 + 20^2)

= sqrt(2500 + 400)

= sqrt(2900)

= 53.852 m/s

(b) angle with the horizontal = tan inverse(50 / 20)

= 68.182 degree

(c) horizontal acceleration = 0

so, horizontal velocity will remain constant

at highest point, horizontal velocity = 20 m/s

the stone will have vertical velocity = 0 when it is at highest point because as the stone goes up, its vertical velocity will decrease and at highest point, it will become instantaneous 0 and will increase in the downward direction..

(d) after 8 seconds, horizonta velocity = 20 m/s

vertical acceleration, a = -g (minus sign to show downwards) = -9.8 m/s^2

initial vertical velocity = u = 50 m/s

v after 8 seconds = u + a * t

v = 50 - 9.8 * 8

= - 28.4 m/s

= 28.4 m/s downwards

resultant = sqrt(20^2 + 28.4^2)

= sqrt(400 + 806.56)

= 34.73 m/s

(e) distance travelled horizontally in that time = horizontal velocity * 8

= 20 * 8 m

= 160 m

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