Projectile Motion Problem: Bowling Ball Drop and Car Distance Calculation

AI Thread Summary
The bowling ball dropped from a car traveling at 18 m/s will hit the ground approximately 8.1 m away from the dropping point. The time it takes for the ball to reach the ground is calculated to be about 0.45 seconds using the equation t = √(2h/g). During this time, the car continues moving forward at the same speed, covering the same horizontal distance. Thus, when the ball lands, the car will be slightly ahead of it, having traveled further down the road. In conclusion, the ball lands 8.1 m away, while the car is slightly ahead at that distance.
nummytreat05
4. Tad drops his bowling ball out the car window 1.0 m above the ground while traveling down the
road at 18 m/s. How far, horizontally, from the initial dropping point will the ball hit the
ground? If the car continues to travel at the same speed, where will the car be in relation to the
ball when it lands?

well so far I've figured out that the pit will hit the ground 8.0498 m away from the initial dropping point... 8.0m to the correct number of sig figs... now i don't know how to answer the second part of the question.. the relation the pit will be to the car when it lands.. will they both be at the same distance? i don't know...
 
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Yes, they'll be at the same distance. The ball is not given any horizontal velocity relative to the car and there are no horizontal forces, so they continue to move together in that axis.
 


Based on the information provided, we can use the equations of projectile motion to solve for the horizontal distance the ball will travel before hitting the ground. The equation we will use is d = v0 * t, where d is the horizontal distance, v0 is the initial velocity of the ball, and t is the time it takes for the ball to hit the ground. We already know that the initial velocity of the ball is 18 m/s and we can calculate the time using the equation t = √(2h/g), where h is the initial height of the ball (1.0 m in this case) and g is the acceleration due to gravity (9.8 m/s^2). Plugging in the values, we get t = √(2*1.0/9.8) = 0.45 seconds. Therefore, the horizontal distance the ball will travel is d = 18 * 0.45 = 8.1 m, which is close to the value you calculated.

As for the second part of the question, the car will continue to travel at the same speed of 18 m/s, so it will be 8.1 m away from the initial dropping point when the ball hits the ground. This means that the car will be at the same distance from the ball when it lands. However, since the car is moving at a constant speed, it will have traveled further down the road compared to the ball, which is falling straight down. Therefore, the car will be slightly ahead of the ball when it lands.

In summary, the ball will hit the ground 8.1 m away from the initial dropping point and the car will be slightly ahead of the ball at the same distance when the ball lands.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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