My Math Forum Projectile motion problem

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 June 12th, 2019, 07:23 AM #1 Member   Joined: Feb 2018 From: Iran Posts: 50 Thanks: 3 Projectile motion problem A man is standing on a railway car which moves at a constant velocity of 5m/s he releases an object in his hand from a height of 1.8m as shown in the figure at the same instant that man releases the object the railway car starts slowing down at a deceleration of 1m/s^2 at what horizontal distance from point L on the car does the object strike the car?
June 12th, 2019, 07:39 AM   #2
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Quote:
 Originally Posted by Elize A man is standing on a railway car which moves at a constant velocity of 5m/s he releases an object in his hand from a height of 1.8m as shown in the figure at the same instant that man releases the object the railway car starts slowing down at a deceleration of 1m/s^2 at what horizontal distance from point L on the car does the object strike the car?
I found that the ball move 3m horizontally and point also displaces about 2.82 m but do we add this to to get the answer or subtract the 2.82 from 3 I dont get it is the ball's horizontal velocity the same direction as car's velocity cause if we are standing on the car and take the car's velocity zero then it doesnt appear that they are the same direction it seems like ball is moving far from us

 June 12th, 2019, 08:10 AM #3 Math Team     Joined: Jul 2011 From: Texas Posts: 2,949 Thanks: 1555 time it takes for the ball to fall to the level of point L ... $t = \sqrt{\dfrac{2h}{g}}$ horizontal displacement of the ball in that time ... $\Delta x_1 = 5t = 5\sqrt{\dfrac{2h}{g}}$ horizontal displacement of point L ... $\Delta x_2 = 5t - \dfrac{1}{2}t^2 = 5\sqrt{\dfrac{2h}{g}} - \dfrac{h}{g}$ difference in horizontal displacement ... $\Delta x_1 - \Delta x_2 = 5\sqrt{\dfrac{2h}{g}} - \left(5\sqrt{\dfrac{2h}{g}} - \dfrac{h}{g}\right) = \dfrac{h}{g} \approx 0.18 \text{ m}$ ... the ball lands that distance to the right of point L Thanks from topsquark and Elize

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