How Far Will the Baby Be from the Platform When Batman Catches Him?

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Homework Statement
One problem with projectile/freefall motion
Relevant Equations
Position equations/I think Center of Mass Equation
Equations with Question
The Batman is attempting to catch the Joker. The Joker stands on a platform near an exit and Batman is on the floor below. Knowing how fast Batman can move up to the platform the Joker uses a ploy to slow the Batman down allowing himself to escape.

Standing on the platform’s edge so that the Batman can see him, the Joker pulls out a 7.00kg baby in a 4.30kg basket, and yells to the Batman and the throws the baby and basket into the air with a speed of 7.00and an angle of 60.00degrees above the horizontal yelling “Congratulations it’sa bouncing baby boy...hopefully”.

The Batman who is standing 7.00m away from the platforms edge of course gives up the pursuit and races to catch the basket. The baby and basket are thrown initially at a height of 6.40m above the floor that the Batman in on. To make meters worse the baby basket is a spring loaded “Joker Basket”and when the Baby and basket at the maximum height the basket launches the baby out directed away from the platform.

Given the catching height of batman, and that the basket launch does not affect the time that the baby is in the air, how fast must Batman run to catch the baby at a height of 1.70m above the ground. Given that the basket when at that height is 3.30m from the platform.

If the speed you find is not humanly possible, it’s okay...this isthe Batman after all.IMPORTANT MATH RULES, for all trig functions and all but the last calculationround to 4 decimal places, your final answer must be rounded to 2 decimal places, and the acceleration due to gravity is 9.8 m/s2 down.
 

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CoachPatrovi said:
Homework Statement:: One problem with projectile/freefall motion
Relevant Equations:: Position equations/I think Center of Mass Equation
Equations with Question

The Batman is attempting to catch the Joker.
The Joker stands on a platform near an exit and Batman is on the floor below. Knowing how fast Batman can move up to the platform the Joker uses a ploy to slow the Batman down allowing himself to escape.

Standing on the platform’s edge so that the Batman can see him, the Joker pulls out a 7.00kg baby in a 4.30kg basket, and yells to the Batman and the throws the baby and basket into the air with a speed of 7.00 and an angle of 60.00 degrees above the horizontal yelling “Congratulations it’s a bouncing baby boy...hopefully”. The Batman who is standing 7.00m away from the platform's edge of course gives up the pursuit and races to catch the basket.

The baby and basket are thrown initially at a height of 6.40m above the floor that the Batman is on. To make matters worse the baby basket is a spring loaded “Joker Basket”and when the Baby and basket at the maximum height the basket launches the baby out directed away from the platform.

Given the catching height of batman, and that the basket launch does not affect the time that the baby is in the air, how fast must Batman run to catch the baby at a height of 1.70m above the ground? Given that the basket when at that height is 3.30m from the platform.

If the speed you find is not humanly possible, it’sokay...this isthe Batmanafter all.
IMPORTANT MATH RULES, for all trig functions and all but the last calculationround to 4 decimal places, your final answer must be rounded to 2 decimal places, and the acceleration due to gravity is 9.8 m/s2 down.
Per forum rules, you need to show some attempt. I see a diagram but no development of equations or discussion of approaches.

There are several things you will need to work out along the way, such as the horizontal velocities before and after the spring acts, and the time for the baby to be at catching height.
Have a go at those.
 
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Delta2 said:
@haruspex how do you understand the eject event from the basket? Does it mean that the velocity of the baby just after the eject is purely horizontal?
Yes I go the initial components which for the horizontal would 3.5 m/s and for the vertical it would be 6.0621m/s
As for the time it would take to reach the catching height of 1.7 meters I got 1.77 seconds. For the Components after the spring I got the basket would be .9797 m/s and for the baby it would be 5.0982 m/s
I also solved for the time it would take to reach max height which was 8.25 meters and it would take .6186 seconds and for it to reach 1.7 meters from 8.25 it would be 1.1584 seconds.
 
Delta2 said:
@haruspex how do you understand the eject event from the basket? Does it mean that the velocity of the baby just after the eject is purely horizontal?
I believe so. My physics teacher words things very strangely so I'm not one hundred.
 
CoachPatrovi said:
Yes I go the initial components which for the horizontal would 3.5 m/s and for the vertical it would be 6.0621m/s
As for the time it would take to reach the catching height of 1.7 meters I got 1.77 seconds. For the Components after the spring I got the basket would be .9797 m/s and for the baby it would be 5.0982 m/s
I also solved for the time it would take to reach max height which was 8.25 meters and it would take .6186 seconds and for it to reach 1.7 meters from 8.25 it would be 1.1584 seconds.
That all looks right. So how far will the baby be from the platform when batman catches him?