Question

How high a human can throw a ball if he can throw it with an intial velocity of 90km/h

Answer 1

The height of a projectile is given by the equation h(t)=1/2 g(t)²+vt+c, where h(t) is the height at t seconds, g is gravity, v is the initial velocity, and c is the projectile’s height at launch. The maximum is determined at time t by the expression t= -b/2a, where b and a are the coefficients of velocity and acceleration (gravity), respectively. So:

90 km/hr=90000/3600=25 m/s initial velocity

-25/ 2 (-4.9)=2.551 secs

h(2.551)=-4.9 (2.551)² + 25(2.551)=-31.8872449+63.775 =31.8877551 m