If you could explain step by step for both A and B, Thanks!! You\'re standing on

Question

If you could explain step by step for both A and B, Thanks!!

You're standing on a cliff overlooking the ocean when you suddenly feel the need to throw a rock over the cliff into the ocean. You calculate the height from your point of view to the water to be H = 225m. You notice that between you and the ocean there is a hump h = 150m in height above the water that you must clear as you throw the rock. If you throw the rock at an angle of theta = 18.5 degree below the horizontal, with what initial velocity v_o must it be thrown in order to overcome/surpass the hump h and reach the ocean? What is the total distance from where you stand on the x-axis to the point the rock enters the water?

Explanation / Answer

Resolve the velocity component along x and y axis.

So Vx = V cos 18.5 = 0.948 V

Simlarly Vy = V sin (18.5) = 0.317 V

Now from equations of motion

x = 0.948 Vt

y = 0.317Vt + 0.5 * g * t^2

So 225 - 150 = 75

So y must be less than 75 to supass and to just graze it should be equal to 75.

So 75 >= 0.317 Vt + 4.9 t^2

Clearly we have one equation and 2 variables. We don't have the time at which it reaches the hump or how far you are standing away from the hump.

Even if we add horizontal direction x = 0.948 Vt we have 2 equations and 3 variables.

4.9t^2 + 0.317Vt -75 <= 0

t^2 + 0.0647Vt - 15.3 <= 0

(t  + 0.03235V)^2 -15.3 - 0.00104V^2 <=0

We need one more ewuation to solve.

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