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Physics

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Step-by-step solution

6 steps

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

    Formula for maximum height: h=u22gh = \frac{u^2}{2g}

    Substitute: h=(20)22×10=40020=20h = \frac{(20)^2}{2 \times 10} = \frac{400}{20} = 20 m

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

    20 m

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  2. Car starts from rest, a=3a = 3 m/s², t=8t = 8 s.

    Find final velocity vv.

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

  3. Distance ss.

    s=ut+12at2=0+12×3×64=96s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 3 \times 64 = 96 m

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

    v = 24 m/s, s = 96 m

    :

  4. Free fall from rest, t=4t = 4 s, g=10g = 10 m/s².

    Find velocity and distance.

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

    s=12gt2=12×10×16=80s = \frac{1}{2}gt^2 = \frac{1}{2} \times 10 \times 16 = 80 m

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

    v = 40 m/s, s = 80 m

    :

  5. Projectile, u=30u = 30 m/s at 37°, g=10g=10 m/s².

    Horizontal range.

    ux=30cos⁡37∘=30×0.8=24u_x = 30 \cos 37^\circ = 30 \times 0.8 = 24 m/s

    uy=30sin⁡37∘=30×0.6=18u_y = 30 \sin 37^\circ = 30 \times 0.6 = 18 m/s

    Time of flight: T=2uyg=2×1810=3.6T = \frac{2u_y}{g} = \frac{2 \times 18}{10} = 3.6 s

    Range R=uxT=24×3.6=86.4R = u_x T = 24 \times 3.6 = 86.4 m

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

    86.4 m

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  6. Deceleration a=−4a = -4 m/s², initial v=50v=50 m/s, find stopping distance.

    Use v2=u2+2asv^2 = u^2 + 2as, v=0v=0

    0=502+2(−4)s0 = 50^2 + 2(-4)s

    200s=2500200 s = 2500

    s=12.5s = 12.5 m

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

    12.5 m

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