robert7248
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You stand on a bridge above a river and drop a rock into the water below from a height of 25 m. (Assume no air resistance)

a. What is the acceleration of the rock as it falls?

b. How long does it take to hit the water?

c. How much more time would it take if instead of dropping the rock straight down you threw horizontally with a speed of 20 m/s?

d. You throw another rock at an angle of 65 degrees above the horizontal with a speed of 20 m/s. What are the x- and y- components of the rock's speed?

e. How long does it take the rock to reach its maximum height?

Respuesta :

PART a)

here when stone is dropped there is only gravitational force on it

so its acceleration is only due to gravity

so we will have

[tex]a = g = 9.8 m/s^2[/tex]

Part b)

Now from kinematics equation we will have

[tex]y = v_i t + \frac{1}{2} at^2[/tex]

now we have

y = 25 m

so from above equation

[tex]25 = 0 + \frac{1}{2}(9.8 )t^2[/tex]

[tex] t = 2.26 s[/tex]

Part c)

If we throw the rock horizontally by speed 20 m/s

then in this case there is no change in the vertical velocity

so it will take same time to reach the water surface as it took initially

So t = 2.26 s

Part D)

Initial speed = 20 m/s

angle of projection = 65 degree

now we have

[tex]v_x = vcos\theta[/tex]

[tex]v_x  = 20 cos65 = 8.45 m/s[/tex]

[tex]v_y = vsin\theta[/tex]

[tex]v_y = 20 sin65 = 18.13 m/s[/tex]

PART E)

when stone will reach to maximum height then we know that its final speed in y direction becomes zero

so here we can use kinematics in Y direction

[tex]v_f - v_y = at[/tex]

[tex]0 - 18.13 = (-9.8) t[/tex]

[tex]t = 1.85 s[/tex]

so it will take 1.85 s to reach the top

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