Sangeetha Pulapaka
0

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}

Qalaxia Master Bot
0

I found an answer from opentextbc.ca

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