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Suppose the stone is thrown at an angle of 39.0° below the horizontal from the same building as in the Example above. If it strikes the ground 47.8 m away, find the following. (Hint: For part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.)

(a) the time of flight
s
The x coordinate as a function of time is x(t) = vcos(39.0)t, so the initial speed is v0 = Δx/(cos 39.0Δt), where Δx = 47.8 and Δt is the time of flight. Insert this into your equation for y(t) and solve for the time of flight.

User Jlmurph
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Suppose the stone is thrown at an angle of 39.0° below the horizontal from the same building as in the Example above. If it strikes the ground 47.8 m away, find the following. (Hint: For part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.)(a) the time of flight sThe x coordinate as a function of time is x(t) = vcos(39.0)t, so the initial speed is v0 = Δx/(cos 39.0Δt), where Δx = 47.8 and Δt is the time of flight. Insert this into your equation for y(t) and solve for the time of flight. Note that the answer should be smaller than 3.16227766016838, since the stone is thrown down (and to the right).(b) the initial speed m/s(c) the speed and angle of the velocity vector with respect to the horizontal at impactspeed m/sangle °
User Flgn
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