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October 31, 2014

October 31, 2014

Posted by **Anonymous** on Monday, August 4, 2014 at 7:37pm.

(a) The maximum height reached by the projectile,

(b) The total time in the air,

(c) The total horizontal distance covered (that is, the range), and

(d) The velocity of the projectile 1.50 s after firing.

- physics -
**Damon**, Monday, August 4, 2014 at 7:57pmdo vertical problem

find Vi, initial speed up

Vi = 65.2 sin 34.5

find t where v = 0, top of curve

v = Vi - 9.81 t

0 = Vi - 9.81 t

solve for t, time at top, half the time in the air

max height

h = 0 + Vi t - 4.9 t^2

now horizontal problem

u = 65.2 * cos 34.5 the whole time

range = u*time in air = u (2 t)

(d)

horizontal component = u the whole time

vertical component = v = Vi - 9.81 (1.5)

speed = sqrt (u^2 + v^2)

tan angle to horizontal = v/u

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