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Initial velocity, u = 30 m/s

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2 parts
a

3 s\text{3 s}

b

45 m\text{45 m}

Step-by-step solution

6 steps

Here are the calculations:

Given:
Initial velocity, u=30 m/su = 30 \, \text{m/s}
Acceleration due to gravity, g=10 m/s2g = 10 \, \text{m/s}^2 (acting downwards)
At the highest point, the final velocity, v=0 m/sv = 0 \, \text{m/s}.

Part a· 3 steps

The time taken to reach its highest point

  1. Use the kinematic equation relating initial velocity, final velocity, acceleration, and time.

    Since the ball is moving upwards against gravity, the acceleration is −g-g.

    v=u−gtv = u - gt

  2. Substitute the known values into the equation.

    0=30 m/s−(10 m/s2)t0 = 30 \, \text{m/s} - (10 \, \text{m/s}^2)t

  3. Solve for tt.

    10t=3010t = 30 t=3010t = \frac{30}{10} t=3 st = 3 \, \text{s}

    The time taken to reach the highest point is 3 s\boxed{\text{3 s}}.

Part b

45 m\text{45 m}

The distance travelled

  1. Use the kinematic equation relating initial velocity, final velocity, acceleration, and displacement.

    v2=u2−2gsv^2 = u^2 - 2gs

  2. Substitute the known values into the equation.

    02=(30 m/s)2−2(10 m/s2)s0^2 = (30 \, \text{m/s})^2 - 2(10 \, \text{m/s}^2)s 0=900 m2/s2−20s m/s20 = 900 \, \text{m}^2/\text{s}^2 - 20s \, \text{m/s}^2

  3. Solve for ss.

    20s=90020s = 900 s=90020s = \frac{900}{20} s=45 ms = 45 \, \text{m}

    The distance travelled to the highest point is 45 m\boxed{\text{45 m}}.

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