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September 1, 2014

September 1, 2014

Posted by **Kelcie** on Thursday, October 15, 2009 at 1:18am.

a. How much higher or lower is the launch point relative to the point where the projectile hits the ground?

b. To what maximum height above the launch point does the projectile rise?

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

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

- Physics -
**MathMate**, Thursday, October 15, 2009 at 8:18amWith projectiles fired at an angle, we consider the vertical component (vy=v0 sin(θ)) and horizontal component (vx=v0 cos(θ)) separately.

vx stays constant throughout.

vy can be treated like a free fall after an iniital velocity of vy.

(a)

So assuming the projectile is fired at y=0, then calculate time t to reach maximum height attained using

t=vy/g

Subtract from total time T of 7.5 seconds to get the time of descent t1.

The distance descended, S1 is given

by

S1 = 0*t1+(1/2)(-g)t1²

Take the difference of S and S1 to get the difference of elevations. Watch the signs.

(b)

use S=(0-vy²)/(2(-g))

(c)

The vertical (downwards) velocity at impact is given by

vy1=t1*(-g)

Add this vectorially to vx to get the velocity.

(d)

the angle θ below the horizontal can be obtained by the arctangent of the ratio of vertical and horizontal components of velocity, namely:

tan(θ) = vy1/vx

(watch the signs, &theta should be between 0 and 90 degrees).

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