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November 23rd, 2016, 05:04 PM   #1
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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
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November 23rd, 2016, 05:42 PM   #2
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$\Delta x$ 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 06:09 PM.
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