A ball is thrown toward a cliff of height h with a speed of 30 m/...

Question

A ball is thrown toward a cliff of height h with a speed of 30 m/s and an angle of 60° above horizontal. It lands on the edge of the cliff 4.0 s later.b. What was the maximum height of the ball?

Accepted Answer

Hey, everyone in this problem, we have some boys playing basketball. One of them is gonna shoot the ball towards the basket, which is at the height of age from the ground. We're told that the ball has a takeoff speed of 12 m per second and a launch angle at 50 degrees above the horizontal. And the ball passes through the basket after three seconds. And we're asked to determine the largest vertical displacement experienced spud ball. We have four answer choices all in meters. Option A 1.34, option B 3.41, option C 4.31 and option D 1.43. Let's do a little diagram of what we have. We have our ball and the ball is gonna be shot towards that net at an angle of 50 degrees with the horizontal with a takeoff speed of 12 m per second. And that ball is gonna travel up, up, up, up, up and then it's gonna come back down before going in the basket. Hm. All right. And we know that that basket has a height of h so the largest vertical displacement that's not gonna be at the basket, that's not the value of age. It's actually gonna be the highest point where the ball reaches before coming back down through the basket. All right. So that's what we're looking for. The largest vertical displacement. And we're gonna call that delta Y. So if we think about this problem, we want to think about our kinematic or U AM equations. OK. So let's write out all the variables we have and what we're trying to find out. OK. And we're gonna write out our variables to the maximum displacement. All right. So if you think about the Y direction, we have the, the initial velocity in the Y direction. Well, that's gonna be related to our angle of 50 degrees through sine because we're talking about the opposite side for the Y component. So we have 12 m per second multiplied by sine of 50 degrees. The final velocity we can say is equal to zero m per second work fall. So when an object reaches its maximum height, its speed is going to be zero and its vertical speed is going to be zero momentarily as it changes from going upwards with positive velocity to downwards with a negative velocity. And so we have this final velocity of zero m per second. And I just mentioned, but we're gonna take up to be positive. All right now, our acceleration, we're gonna have the acceleration due to gravity which is negative 9. m per second squared. And now we have delta Y, which is what we are looking for. That's that he, we drew in red on our diagram. We aren't given any information about the time it takes to get to the maximum displacement and we are given the time it takes to complete a basket. So we have that if we need it. But if we look at what we have, we have three known values V not Y V F Y and A, we wanna find delta Y, OK. With three known values we can solve for the fourth, we just need to choose the correct equation. We're gonna choose the equation that has our three known values as well as delta Y, which is what we're looking for. And that equation is gonna be V F Y squared is equal to V not right word plus two A delta Y OK. Now we have our equation. We have all our variables. All that's left to do is substitute them in and solve for delta Y. On the left hand side, we get zero. This is gonna be equal to V, not Y squared. So 12 m per second multiplied by sine of 50 degrees squared plus two, multiplied by the acceleration, negative 9.8 m per second, scored multiplied by delta Y that horizontal displacement or sorry, the vertical displacement that we're looking for. All right. So we want to solve for delta Y. We're gonna do that just like we would solve any equation. We're gonna move our two multiplied by 9.8 m per second squared, multiplied by delta Y term to the left hand side. We're gonna simplify. So we have 19.6 m per second squared multiplied by delta Y it's equal to 12 m per second, multiplied by a sine of 50 degrees all squared. We're just gonna leave that like that for now till the very end. So we can avoid some round off error here. In order to isolate for delta Y, we're gonna divide by 19.6 m per second squared. Yet delta Y is equal to 12 m per second, multiplied by sine of 50 degrees all squared divided by 19.6 m per second squared. In terms of our units, we have meters per second all squared in the numerator which gives us meters squared per second squared. And then we're dividing by meters per second squared. So all that's gonna be left is a unit of meter, which is what we want for this vertical displacement. And we get the delta Y is approximately equal to 4.31 m. All right. So looking at our answer choices, we found that the maximum vertical displacement experienced by the basketball is 4.31 m which corresponds with answer choice. C thanks everyone for watching. I hope this video helped see you in the next one.

Struggling with Physics?

A ball is thrown toward a cliff of height h with a speed of 30 m/s and an angle of 60° above horizontal. It lands on the edge of the cliff 4.0 s later.b. What was the maximum height of the ball?

Watch next

Positive (Upward) Launch practice set