NOTE: Take UP as POSITIVE. A projectile is shot on level ground with a horizonta
ID: 1762815 • Letter: N
Question
NOTE: Take UP as POSITIVE.A projectile is shot on level ground with a
horizontal velocity of 29 m/s and a vertical
velocity of 66 m/s .
The acceleration of gravity is 9.8 m/s2 .
a) Find the travel time of the projectile total.
Answer in units of s. (I already found it : 3.688s)
b) Find the range R. Answer in units of m. (I already foundit: 8.663 m)
c) Now consider a new situation, where the
magnitude of v0 is fixed to be the same value as
that in part 1 and part 2, and we vary the
angle to get the maximum range.
Find the height when the range is maximum. Answer in unitsof m. (Question)
Explanation / Answer
vx = 29 m/s, vy = 66 m/s a) vertical displacement = vyt - gt2/2 =0 t = 2vy/g = 13.47 s b) R = vxt = 390.6 m c) v0 =(vx2 + vy2) =72.09 m/s when the range is maximum, =45o height = (v0sin)2/(2g) = 132.6 m
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