Answer
part a is 112 part b is 289
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<span>A projectile is fired with an initial velocity of 450 ft / sec and with an elevation angle of 30 ° above the horizontal, calculate
a) its position and speed after 8s,
The position after 8 seconds will be as follows
horizontal distance:
x=V</span>×t
x=450×8
=3600 ft
b) the maximum height reached will be given by:
H=[v² sin²θ]/2g
Plugging in the values we get:
H=[450²×sin²30]/(2×10)
H=2,531.25 ft
Answer:
7/500
Step-by-step explanation:
1/10*1/5*7/10