You throw a glob of putty straight up toward the ceiling, which is 3.60 m above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 9.50 m/s. (a) What is the speed of the putty just before it strikes the ceiling?

Video transcript

Hello everyone. So in this problem you throw a cricket ball vertically upward with a speed of 10 m per second towards the rooftop which is four m above the point where the ball use your hand. What is the speed of the cricket ball just before it hits the rooftop? So to conceptualize this problem, we have some ball Which is thrown with a speed of 10 m/s. Some rooftop four m, both the point, So four m represents the change in height. We have some initial velocity and what we directly want to find. It's the velocity When it reaches that four m difference. So we recall our list of cinematic equations. B. F equals D. I plus the F squared equals the I squared. Close to a W. X. And delta X. V I. T plus one half 18 square. Now in this problem we noticed that we have V. I and delta H. And we're specifically looking for the we also recall that since this ball is in free throw it's A. Is equal to the acceleration due to gravity just negative 9. meters per second square. So we also have A or G. Among these equations, we can see that the best equation to use which has all this information is equation too. So we can start out by writing the equation to this problem. B f squared equals B I squared Plus two G. and Delta X is simply just delta H. Just distance. Now we can directly solve for B F. Just taking the square root on the other side. Now we can plug in numbers the F to the square root per second square times nine point here is swear delta H is for This equation now gives us a value for b. of 4.65 m/s. Answer choice of the sauce Have a great day.