Physics Help Forum Trajectory of projectile that clears obstruction

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Jan 26th 2018, 05:00 PM #1 Junior Member   Join Date: Jan 2018 Posts: 1 Trajectory of projectile that clears obstruction My problem is I need to solve for the minimum Vx and minimum Vy so that the projectile clears the gray wall and still hits its mark defined by (x1 + x2). This would be the red trajectory line. I am aware of how to solve the blue trajectory line. The blue line exists in this diagram only to show the difference in scenarios (I am not concerned with the blue line). My givens are: x1 x2 y1 y2 y3 (which is the target) is always 0 PLEASE NOTE: this is not your typical textbook "throw the ball over the house" problem. It differs in that x1 and x2 are not equal AND y2 is not the maximum height of the trajectory. How do I go about solving this?? Any ideas?
 Jan 27th 2018, 08:41 AM #2 Senior Member   Join Date: Aug 2010 Posts: 282 The only acceleration here is the acceleration due to gravity, -g. There is no horizontal acceleration so the horizontal velocity is the constant, $v_x$. After time t, the horizontal distance is $x(t)= v_xt$. With constant acceleration -g, the height at time t is give by $y(t)= -(g/2)t^2+ v_yt+ y_1$. y will be 0 when $y(t)= -(g/2)t^2+ v_yt+ y_1= 0$. There are two solutions to that quadratic equation- you want the positive one. Setting $x(t)= v_xt= x_1+ x_2$ for that t gives one equation for $v_x$ and $v_y$. The wall is distance $x_1$ from the initial position so when the projectile is passing the wall, $v_xt= x_1$, $t= \frac{x_1}{v_x}$. At that time, the height of the projectile will be $y= -(g/2)\frac{x_1^2}{v_x^2}+ \frac{x_1v_y}{v_x}+ y_1$ and we want that to be larger than $y_2$. That gives a second equation so you have two equations to solve for $v_x$ and $v_y$.

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