Extract the homework question from this image. State the question concisely (under 200 chars). Identify the academic subject.

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Step 1: Given initial velocity $u = 20$ m/s upward, $g = 10$ m/s² downward, find maximum height.

Formula for maximum height: $h = \frac{u^2}{2g}$

Substitute: $h = \frac{(20)^2}{2 \times 10} = \frac{400}{20} = 20$ m

Final answer: $h = \mathbf{20\ m}$

20 m

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Step 1: Car starts from rest, $a = 3$ m/s², $t = 8$ s. Find final velocity $v$.

$v = u + at = 0 + 3 \times 8 = 24$ m/s

Step 2: Distance $s$.

$s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 3 \times 64 = 96$ m

Final answer: $v = \mathbf{24\ m/s}$, $s = \mathbf{96\ m}$

v = 24 m/s, s = 96 m

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Step 1: Free fall from rest, $t = 4$ s, $g = 10$ m/s². Find velocity and distance.

$v = gt = 10 \times 4 = 40$ m/s

$s = \frac{1}{2}gt^2 = \frac{1}{2} \times 10 \times 16 = 80$ m

Final answer: $v = \mathbf{40\ m/s}$, $s = \mathbf{80\ m}$

v = 40 m/s, s = 80 m

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Step 1: Projectile, $u = 30$ m/s at 37°, $g=10$ m/s². Horizontal range.

$u_x = 30 \cos 37^\circ = 30 \times 0.8 = 24$ m/s

$u_y = 30 \sin 37^\circ = 30 \times 0.6 = 18$ m/s

Time of flight: $T = \frac{2u_y}{g} = \frac{2 \times 18}{10} = 3.6$ s

Range $R = u_x T = 24 \times 3.6 = 86.4$ m

Final answer: $R = \mathbf{86.4\ m}$

86.4 m

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Step 1: Deceleration $a = -4$ m/s², initial $v=50$ m/s, find stopping distance.

Use $v^2 = u^2 + 2as$, $v=0$

$0 = 50^2 + 2(-4)s$

$200 s = 2500$

$s = 12.5$ m

Final answer: $s = \mathbf{12.5\ m}$

12.5 m