Hi everyone! I\'m a little confused as how to start thisproblem. I\'m okay with

Question

Hi everyone! I'm a little confused as how to start thisproblem. I'm okay with solving some problems, but the wordingis confusing me.

Question - A projectile is shot at an angle of 45 degreesto the horizontal near the surface of the earthbut in the absence of air resistance. When it reaches thehighest point of its trajectory, its speed is 150 m/s. In asecond trial with the same projectile, the initial speed is thesame but the angle is now 37 degrees with the horizontal. Atits highest point its trajectory, the speed of the projectile wouldbe?

a)150 m/s (sin45 degrees/sin36 degrees)
b)150 m/s (cos37 degrees/cos45 degrees)
c)150 m/s (sin37 degrees/sin45 degrees)
d)150 m/s (37/45)
e) none

Thanks so much! The bold is what Im having troublewith....it's hard to picture in my mind and on paper.

Explanation / Answer

at the highest point the projectile's vertical velocity=0 it's horizontal velocity=150 m/s let the initial velocity of the projectile=u m/s initial horizontal velocity=u*cos45 horizontal acceleration=0 v=u-a*t 150=u*cos45+0 u=150/cos45 now angle of projectile=37 degree horizontal velocity=(150*cos37)/cos45 since the horizontal acceleration=0 velocity at the highest point=(150*cos37)/cos45 so the answer=b)(150*cos37)/cos45

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