The equation of motion with constant acceleration is given by,

v^{2} = u^{2} + 2as

The initial velocity, u, is 1.6 m/s, the acceleration due to gravity, a = -9.81\frac{m}{s^{2}}

The displacement is negative at the highest point , so s = -3.0 m

Substituting these in the equation we get

v^{2} = \Big(1.6 \frac{m}{s}\Big)^{2} + 2 \times \Big(-9.8 \frac{m}{s^{2}}\Big) \times (-3.0 m)

v^{2} = 2.56 \frac{m^{2}}{s^{2}} + 58.8 \frac{m^{2}}{s^{2}}

v^{2} = 61.36 \frac{m^{2}}{s^{2}}

v = 7.833 \frac{m}{s}

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3.5 Free Fall – University Physics Volume 1

**Acceleration** due to gravity is **constant**, which means we can apply the **kinematic equations** to any falling object where air resistance and friction are ...

For more information, see 3.5 Free Fall – University Physics Volume 1