Ask your homework questions to teachers and professors, meet other students, and be entered to win $600 or an Xbox Series X 🎉Join our Discord!

Like

Report

Carnegie Mellon University

Like

Report

Problem 17 Easy Difficulty

A projectile is launched with an initial speed of 60.0 $\mathrm{m} / \mathrm{s}$ at an angle of $30.0^{\circ}$ above the horizontal. The projectile lands on a hillside 4.00 s later. Neglect air friction. (a) What is the projectile's velocity at the highest point of its trajectory? (b) What is the straight-line distance from where the projectile was launched to where it hits its target?

Answer

(a) $v _ { 0 x } = 51.96 \mathrm { ms } ^ { - 1 }$
(b) $s = 211.96 \mathrm { m }$

Discussion

You must be signed in to discuss.

Video Transcript

the highest point of the trajectory. The projectile is actually moving perfectly horizontally with no, um why component of the velocity. And so we can say that four part, eh? Our velocity in next direction will be equaling the the initial velocity in the extraction. Given that there isn't any acceleration in the ex direction, this would be equaling the initial velocity multiplied by co sign of data. And so this would be equaling 60 meters per second multiplied by co sign of 30 degrees. And this is giving us 52.0 meters per second. Now, this would be our answer for Part II for part B. Then we know that the horizontal displacement delta axes equaling the velocity and the extraction times t so this would be equaling 52 0.0 meters per second multiplied by 4.0 seconds. And this is giving us 208 meters. And at this point, we can say that delta y equaling the initial sign of data that would beer initial velocity in the UAE direction multiplied by t and then plus 1/2 times the acceleration the Y direction times t squared Times Square. So the vertical displacement delta y is going to be equaling 60 meters per second time sign of 30 degrees multiplied by 4.0 seconds plus 1/2 multiplied by negative 9.80 meters per second squared multiplied by 4.0 seconds. Quantity squared and this is equaling 41.6 meters. So we can then say that the straight line distance for part B the straight line, we can simply say D for the displacement. This would be equal in the square root of Delta X quantity squared plus delta y quantity squared. Extend the square root. This is gonna be equal in the square root of 208 meters quantity squared plus 41.6 meters quantity squared and we find that d the magnitude of the straight line displacement is equaling. 212 meters. This would be our final answer for part B. That is the end of the solution. Thank you for watching

Carnegie Mellon University
Top Physics 101 Mechanics Educators
Elyse G.

Cornell University

Marshall S.

University of Washington

Zachary M.

Hope College

Aspen F.

University of Sheffield