SoRry But I Have An a PrOblem To start WIththis.... A boy is at a distance d = 6

Question

SoRry But I Have An a PrOblem To start WIththis....

A boy is at a distance d = 6m from the base of a building attemptsto throw a ball through a window size that H = 90cm at a height h =8m.
if the ball's initial speed is v0 = 15m / s,
determine the range of initial angles 0 allowing the ballpasses through the window. (acceleration of the ball is 9.81 m / s^2 vertical down)

How Can I DetErmine the Angles??

[IMG]http://img411.imageshack.us/img411/5928/123gs.jpg[/IMG]

Its from the Text: Engineering Mechanics: dynamic (Riley &sturges) 1st edition .. 13.89

ThanKs For The help!!

Explanation / Answer

To solve this problem, we use the kinematic equations for constantacceleration. Remembering that the motion involves two separatecomponents we have: y = y0 + vyot - 1/2ayt2 x = x0 + vx0t - 1/2axt2 We also need the equations for the initial velocitiesvx0 and vy0 vx0 = v0 cos() vy0 = v0 sin() Where is the launch angle. Now, you did not give the boy's height so I'm assuming he'll throwit from the ground. That gives: y = y0 + vyot - 1/2ayt2 = vyot - 1/2ayt2 Now, substituting for vy0 we have: y = (sin()v0) t - 4.905t2                    Equation (1) STEP 1: GET THE MINIMUM ANGLE So now we have one equation, but two unknowns, t and (we want y to be 8 meters). BUT, we know that this is aquadratic equation, which makes a parabola, and the equation forthe x (or in this case t) coordinate of the vertex of a parabola is(-b / 2a), where a is the coefficient of the quadratic term and bis the coefficient of the linear term. In this case, a = -4.905 andb = (sin() v0). So -b / 2a = (-sin() v0) / -9.81 =(sin() v0) / 9.81 So, the t at which the ball will reach maximum height (8 feet) isequal to (sin() v0) / 9.81. Therefore, we plugthat in to what we called Equation (1) above toget: y = (sin()v0) * ((sin() v0)/ 9.81) - 4.905 ((sin() v0) /9.81)2                                    Equation (2) Now, we know y (8 m), we know v0 (15 m/s). So, we haveone equation, and one unknown, which can be solved for . Thisis your minimum angle for the range of angles. STEP 2: GET THE MAXIMUM ANGLE Now, the process for getting the maximum angle is exactly the samefor the minimum angle, BUT you must use y= 8.9 meters when solving Equation(2). I hope this helps. Good luck, and don't forget to rate!

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