November 23rd, 2016, 04:04 PM  #1 
Member Joined: Sep 2016 From: zambia Posts: 31 Thanks: 0  Projectile motion
HOW DO YOU GO ABOUT SOLVING THE FOLLOWING QUESTION? A person initially at rest throws a ball upward at an angle !0 with an initial speed v0. He tries to catch up to the ball by accelerating with a constant acceleration a for a time interval !t1 and then continues to run at a constant speed for a time interval !t2. He catches the ball at exactly the same height he threw the ball. Let g be the gravitational constant. What was the person’s acceleration a ? I have only been able to come up with the x and y positions of the ball and I have also shown that initial speed of the ball and angle are related...Now my problem is solving for the horizontal motion of the person 
November 23rd, 2016, 04:42 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,737 Thanks: 1387 
$\Delta x$ and the overall time interval, $t$, will be the same for the thrower and the ball for the projectile ... $\Delta x = v_0\cos{\theta} \cdot t$ $t = \dfrac{2v_0 \sin{\theta}}{g} \implies \Delta x = \dfrac{v_0^2\sin(2\theta)}{g}$ for the thrower, $t_1$ is the time he accelerates and $t_2$ is the time he runs at a constant speed ... $\Delta x = \dfrac{1}{2}at_1^2 + (a t_1) \cdot t_2 = a\left(\dfrac{t_1^2}{2} + t_1 \cdot t_2\right)$ set $\Delta x$ for the projectile equal to $\Delta x$ for the thrower and solve for $a$ ... Last edited by skeeter; November 23rd, 2016 at 05:09 PM. 

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