Reaches a maximum height of 1500m before landing on the...

reaches a maximum height of 150.0m before landing on the ground. The total time of flight for the rocket is 10.0 seconds. To find the horizontal distance traveled by the rocket, we can use the formula: horizontal distance = initial velocity * time * cosine(angle) Using the given values, the horizontal distance can be calculated as: horizontal distance = 76.0m/s * 10.0s * cosine(55.0°) To find the maximum height reached by the rocket, we can use the formula: maximum height = (initial velocity * sin(angle))^2 / (2 * acceleration due to gravity) Using the given values, the maximum height can be calculated as: maximum height = (76.0m/s * sin(55.0°))^2 / (2 * 9.8m/s^2) Please note that the calculations provided are based on the given information and may not be accurate in real-world scenarios.

Transcript

00:01 Hi there.

00:01 So for this problem, we are given the initial speed of this rocket that we're going to label as b -0, and that is 1 ,100 meters per second.

00:13 And this, at an angle, let's label this angle as theta that is equal to 52 degrees.

00:19 So we need to determine the following.

00:21 So for part a, let's level this part a of this problem, the maximum height that the rocket will attain.

00:28 So first of all, we consider the following expression from kinematics that states that the final white component of the speed to the square is equal to the initial white component of the speed to the square, and this minus two times, yes, two times the acceleration, in this case, corresponds to the acceleration due to gravity, and this times the maximum height.

00:58 So we know that at the maximum height, the y component of speed is momentarily equals to zero.

01:06 So what we can do is to just simply solve for that maximum height.

01:11 So that will be the initial white component of the speed to the square, this divided by two times the acceleration due to gravity.

01:18 So that will be, let me simply move this a little bit to the left.

01:24 So that will be the initial way component of the speed to the square, this times the sine square of the angle theta, and this divided by two times the acceleration due to gravity.

01:33 And now we substitute the values in here, so that will be 1 ,100 and that to the square, this times the sine square of the angle theta, which is equal to 52 degrees, and then this divided by two times the acceleration due to gravity, which is 2 times 9 .8.

01:54 Then using our calculator, we obtain a value of.

02:05 So let me just use the calculator.

02:16 So the value that we obtain for this is equal to a maximum height of 38 ,335 meters.

02:37 So that's the solution for the first question for this problem, the maximum height.

02:43 Now, for part b, we are asked about the time that we'll take for the rocket to reach its maximum height.

02:53 Okay, so in this case, we need to use kinematics, okay? so we use the following.

03:03 That is that the, well, we know that the final white component of the speed is equal to the initial white component of the speed in this minus the acceleration due to gravity times the time.

03:24 Now we know that at the maximum height, this final speed is equal to zero.

03:31 So what we need to do is to solve for the time.

03:34 So that will be the initial white component of the speed, which is the initial speed times the sign of the angle that we are given for this.

03:41 The angle theta, this divided by the acceleration due to gravity.

03:44 So now we substitute the values in here.

03:48 So that will be the initial speed that is equal to 1 ,100.

03:55 And this times the sign of the angle that is 52 degrees.

04:02 Okay, 52...