"A group of students launch a projectile straight up at 12 m/s. They record maximum height and total time in the air The students then launch another projectile at an angle of 30 degrees. What launch velocity should they use to produce the same maximum height and same total time in the air?"

they should launch at a velocity of 24 m/s to produce the same maximum height and same total time in the air

Explanation:

If a projectile is launched upward; initial velocity u = 12 m/s

final velocity v = 0 m/s

a = -g = -9.8 m/s²

now from the equation of motion

V = u + at

we substitute

0 = 12 + ( -9.8 × t)

9.8t = 12

t = 12/9.8

t = 1.22 sec

to get maximum height, we say;

H = ut + 1/2at²

we substitute

H = (12 × 1.22) + 1/2( -9.8 × (1.22)²)

H = 14.64 + ( - 7.29 )

H = 14.64 - 7.29

H = 7.35 m

total time in the Air

T = 2 × t

= 2 × 1.22

= 2.44 sec

Now if the projectile is at 30°, the initial velocity u' is ?

given that H = 7.35 m, Tt = 2.44 sec

so

Maximum height H = ( u'²sin²30 / 2g)

u'² = H2g / sin²30

we substitute

u'² = (7.35 × 2( 9.8)) / ( sin²30)

u'² = 144.06 / 0.25

u'² = 576.24

u' = √576.24

u' = 24.0049 ≈ 24 m/s

therefore they should launch at a velocity of 24 m/s to produce the same maximum height and same total time in the air

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