A projectile is fired with an initial speed of 27.0 m/s at an angle of 65.0 abov

Question

A projectile is fired with an initial speed of 27.0 m/s at an angle of 65.0 above the horizontal. The object hits the ground 10.0 s later.

1. How much higher or lower is the launch point relative to the point where the projectile hits the ground? Express a launch point that is lower than the point where the projectile hits the ground as a negative number.

2. What is the magnitude of the projectile's velocity at the instant it hits the ground?

3. What is the direction (below +x) of the projectile's velocity at the instant it hits the ground?

Explanation / Answer

1.
In vertical direction:
Vi = 27* sin 65 = 24.5 m/s
a = -9.8m/s^2
t = 10 s
use:
h = vi*t + 0.5*a*t^2
= 24.5*10 + 0.5*(-9.8) * (10)^2
= - 245 m
It will hit 245 m lower
Ans :- 245 m

2.
Horizontal velocity will be constant always = 27 * cos 65 = 11.4 m/s
consider vertical motio:
Vf = Vi + a*t
=24.5 - 9.8*10
= -73.5 m/s
Magnitude of final velocity = sqrt (11.4^2 + 73.5^2) = 74.4 m/s

3.
angle below x axis = atan (73.5/11.4) = atan (6.45) = 81.2 degree

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