Reviewing what I got wrong, I\'d like to review my new answers with some help th

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Reviewing what I got wrong, I'd like to review my new answers with some help thanks!

2. (20 points) A cannonball is shot at 60' angle above the horizontal, to the right, with a speed of , and 10m above 48.765m/s. The cannonball lands on top of a tall building 200m away horizontally where it was fired from. Set the origin at the point where the cannonball leaves th acceleration vector is 10m/s pointing down. Up and right are positive. a. (12 points) Carefully ill in the graphs of the x and y components of postion, velocity, and acceleration, along with any known values. The vertical line is when the cannonball reaches its highest point. x-direction (rights pos) y-direction (up pos) ay b. (4 points) What is the maximum height of the cannonball, above the height where it was fired? Show your work clearly. Use the back if you need more space (and write that you did so) 2 ( 1357, 71 Sin tbo*) |1.37m (4 points) How long does it take to hit the building? Show your work clearly. Use the back if you need more space (and write that you did so). c. 767

Explanation / Answer

Given

Initial velocity (v) = 48.765 m/s

angle of launch () = 60 deg

horizontal distance of building (R) = 200 m

Height of the building (h) = 10 m

a) x t graph is straight line with a constant slope.as velocity is constant along x axis

Y t graph is a parabola as there is a constant acceleration along y axis.

ax t graph is 0 as acceleration along x axis is 0.

ay is straingt line parallel to t as acceleration is constant along y axis.

b ) Horizontal Range = v^2 sin^2 / (2g) = 48.765^2 * (sin 60 )^2 / (2*10) = 89.17 m

Maximum Height = v^2 sin 2 /g = 48.765^2 * sin 120 / 10 = 205.94 m

c) It will hit the building when the ball covers horizonntal distance of 200 m but the horizontal range of the projectile is only 89.17 m. It will not be able to reach the building and touch the gorund before raching the building.

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