As shown in the figure, for an object launched at height y0 (m), initial velocity V (m/s), and angle θ (degrees) made with the horizontal, when the maximum height ht (m) and horizontal moving distance L (m) are given, V and θ can be calculated. The vertical velocity Vy of the object is 0 at ht, the vertical acceleration is -g, and the time t1(s) taken from y0 to ht is equal to the time taken from ht to y0 by free fall. ht-y0 = (1/2)g*t1^2 t1 = sqr(2*(ht-y0)/g) The time to reach from ht to the bottom is, t2 = sqr(2*ht/g) Thus, ts = t1+t2 = sqr(2*(ht-y0)/g) + sqr(2*ht/g) Horizontal velocity Vx and Vy can be obtained from ts, t1 Vx = L/ts Vy = g*t1 so, initial velocity V is V = sqr(Vx^2 + Vy^2) and the angle made with horizontal is θ = arctan(Vy/Vx)

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