The final velocity of the projectile when it strikes the ground below is 198.51 m/s.
<h3>
Time of motion of the projectile</h3>
The time taken for the projectile to fall to the ground is calculated as follows;
h = vt + ¹/₂gt²
where;
- h is height of the cliff
- v is velocity
- t is time of motion
265 = (185 x sin45)t + (0.5)(9.8)t²
265 = 130.8t + 4.9t²
4.9t² + 130.8t - 265 = 0
solve the quadratic equation using formula method,
t = 1.89 s
<h3>Final velocity of the projectile</h3>
vyf = vyi + gt
where;
- vyf is the final vertical velocity
- vyi is initial vertical velocity
vyf = (185 x sin45) + (9.8 x 1.89)
vyf = 149.322 m/s
vxf = vxi
where;
- vxf is the final horizontal velocity
- vxi is the initial horizontal velocity
vxf = 185 x cos(45)
vxf = 130.8 m/s
vf = √(vyf² + vxf²)
where;
- vf is the speed of the projectile when it strikes the ground below
vf = √(149.322² + 130.8²)
vf = 198.51 m/s
Learn more about final velocity here: brainly.com/question/6504879
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Answer:
As <em>Force </em><em>=</em><em> </em><em>Mass </em><em>×</em><em> </em><em>Acceleration.</em><em> </em><em>So </em><em>If </em><em>mass </em><em>is </em><em>constant,</em><em> </em><em>the </em><em>acceleration </em><em>would </em><em>also </em><em>be </em><em>constant,</em><em> </em><em>in </em><em>presence </em><em>of </em><em>a </em><em>net </em><em>external </em><em>constant </em><em>force!</em>