Question

A diver of $50kg$ jumps from a platform $20m$ high into a pool. If the diver decelerates at a constant rate to zero velocity in $0.8seconds$ after hitting the water, what is the force that the water exerts on the diver? $(g=10m{s}^{-2})$

• A. $1680N$
• B. $850N$
• C. $425N$
• D. $1250N$

Answer 1

The correct option is D $1250N$

Height, $h=20m$
From third equation of motion:
$v^2=u^2+2as$
$u=0,a=10m{s}^{-2},s=h=20m$
$v^2=2\times 10m{s}^{-2}\times 20m$
$v=\sqrt{400}m{s}^{-1}$
$v=20m{s}^{-1}$(Since $v\ne -20m{s}^{-1}$)
We know that the diver then decelerates from this velocity to zero in 0.8 seconds, so we can calculate the acceleration:
$a=\frac{v_f-v_i}{t}=\frac{0-20}{0.8}=25m{s}^{-2}$
(Negative sign denotes that the person is undergoing retardation)
Use Newton's second law to calculate the force on the diver:
$F=ma$
=>$\left(50kg\right)(-25m{s}^{-2})=-1250N$
(Negative sign denotes that the force exerted by water is upwards)