What’s the distance at maximum height?

if α = 0°, then vertical velocity is equal to 0 (Vy = 0), and that's the case of horizontal projectile motion. As sine of 0° is 0, then the second part of the equation disappears, and we obtain : hmax = h - initial height from which we're launching the object is the maximum height in projectile motion.

Step-by-step explanation: but also The range of an object, given the initial launch angle and initial velocity is found with: R=v2isin2θig R = v i 2 sin ⁡ 2 θ i g . The maximum height of an object, given the initial launch angle and initial velocity is found with:h=v2isin2θi2g h = v i 2 sin 2 ⁡ θ i 2 g .

$4

step-by-step explanation:

$30×5=$150

$150 minus $130=$20

$20÷5=$4.

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