2D Motion and Calculus

Question

A projectile is fired from the ground (you can assume the initialheight is the same as the ground) in a field so there are noobstacles in its way. It is fired at an angle 34° with respectto the horizontal and with an initial speed of 47 m/s. Airresistance is negligible in this situation.


What is the maximum height that the projectile reaches?

What is thespeed of the projectile at its maximum height?

The same projectile was then fired in the same way toward a wallthat is a horizontal distance 164.9 m from where theprojectile was fired.
What was the height of the projectile when it hit the wall?



What was the velocity of the projectile when it hit thewall?
Call down the positive y direction, andtoward the wall thepositive x direction.
i +  j;


What was the speed of the projectile when it hit the wall?



If the wall were slightly farther from where the projectile waslaunched (the projectile still hits it), how would that effect thefollowing ( CHOOSE BETWEEN INCREASES, DECREASES OR STAYS THE SAME?
IncreasesDecreasesStays the same The magnitude of the x componentof the velocity of the projectile when it hits the wall.
IncreasesDecreasesStays the same The maximum height of the projectile.
IncreasesDecreasesStays the same The speed of the projectile when it hits thewall.
IncreasesDecreasesStays the same The magnitude of the y componentof the velocity of the projectile when it hits the wall.
IncreasesDecreasesStays the same The distance from the ground to where the projectilehit the wall.

Explanation / Answer

You can break down the velocities into horizontal and verticalcomponents using trig functions... Horizontal velocity = 47cos(34) = 38.96 m/s Vertical velocity = 47sin(34) = 26.28 m/s -- First you have to find the time that it took forthe projectile to reach its max height... vf = vi + at (Note: a is theacceleration of gravity, which is -9.8 m/s/s. Also, vf isfinal velocity and vi is initial velocity. Also, when theprojectile reaches its max height, its velocity will be 0) 0 = 26.28 - 9.8t t = 2.68 seconds -- To find the max height, use this equation and the verticalvelocity... d = vt + 0.5at2 (Note: a is the acceleration ofgravity, which is -9.8 m/s/s) d = (26.28)(2.68) + 0.5(-9.8)(2.68)2 **d = 35.24 m** -- Like we said before, the speed at the max height will be 0because at that point it will have to start moving in the oppositedirection. It will continue moving 38.96 m/s horizontally,though. -- Find the time it takes to go 169.4 m horizontally by using t =d/v = 169.4/(38.96) = 4.34 s Then, find out how high it will be vertically by using the "d"equation... d = (26.28)(4.34) + (0.5)(-9.8)(4.34)2 **d = 21.76 m** --
The horizontal velocity will always be the same, the veritcalvelocity can be found by v = 26.28 - 9.8(4.34) = -16.25 So... **-16.25i + 38.96j** **d = 21.76 m** --
The horizontal velocity will always be the same, the veritcalvelocity can be found by v = 26.28 - 9.8(4.34) = -16.25 So... **-16.25i + 38.96j**

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