Trajectory, where it will land.

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SUMMARY

The projectile is launched from a height of 125 meters with an initial speed of 105 m/s at a 37-degree angle. The total time taken for the projectile to hit the ground is calculated to be 14.63 seconds. The horizontal range from the base of the cliff is determined to be approximately 1226.87 meters. The horizontal and vertical components of the initial velocity are 83.86 m/s and 63.19 m/s, respectively, with the magnitude of the velocity being 105 m/s.

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  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Knowledge of trigonometric functions
  • Basic grasp of gravitational acceleration (9.8 m/s²)
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  • Study the derivation of the range formula R=(Vo^2sin2(theta))/g
  • Learn about the effects of air resistance on projectile motion
  • Explore the concept of maximum height in projectile motion
  • Investigate the use of simulation tools for projectile trajectories
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Homework Statement


a projectile is shot from the edge of a cliff 125m above ground level with an initial speed of 105m/s at an angle of 37 degrees with the horizontal axis.
1.determine the time taken by the projectile to hit ground.
2.determine the range X of the projectile as measured from the base of the cliff.
3.find the horizontal and vertical components of its velocity.
4.find the magnitude of the velocity
5.the angle made by the velocity vector with the horizontal

Homework Equations


R=(Vo^2sin2(theta))/g

The Attempt at a Solution


1.
Vf=Vo+at
0=63.19m/s+-9.8m/s^2t
-63.19m/s=-9.8m/s^2t
t=6.45s that's the time for the projectile going up the Y axis

y=125+ 63.19m/s(6.45s)+-4.9(6.45s)^2
y= 328m

d=1/2(a)(t)^2
328 = 4.9t^2
t^2=66.94
t=8.18s time going down the Y axis

time total=8.18s+6.45s=14.63s

2.
Dx=v*t
Dx=83.86m/s*14.63s
Dx=1226.87m

3. sin(37)(105m/s)=63.19m/s y
cos(37)(105m/s)=83.86m/s x
4.
5.

as you can tell, I need a lot of help in completing this problem.
 
Last edited:
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http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html

Look for "Where will it land?"

Look at the equations of motion for x(t) and y(t).

vx is constant if one neglects air resistance, but vy will vary in time (or height) because of the deceleration due to gravity and the acceleration once the projectile reaches maximum height.

What are v0x and v0y in terms of v0 and the angle with respect to the horizontal axis?
 

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