The acceleration due to gravity practically does not change near the earth and therefore it is considered constant. If the ball is thrown not high, then it will also accelerate $9.8 \frac{m}{s^2}$ (even astronauts on the ISS are affected by the acceleration of gravity of about $9.2 \frac{m}{s^2}$). How to determine the height to which the ball will rise: equate the kinetic energy at the moment of the throw to potential energy at the moment of the highest point. If you raise the ball vertically upward at speed $v$, then $m*\frac{v^2}{2} = mgh$, hence $h = \frac{v^2}{g}$, for example, if you throw the ball at speed $60 \frac{mile}{h} = 96.5 \frac{km}{h} = 26.8 \frac{meter}{s}$ then the ball will take off to a height of $73 meter = 76.5 yards$

The acceleration due to gravity practically does not change near the earth and therefore it is considered constant. If the ball is thrown not high, then it will also accelerate $9.8 \frac{m}{s^2}$ (even astronauts on the ISS are affected by the acceleration of gravity of about $9.2 \frac{m}{s^2}$). How to determine the height to which the ball will rise: equate the kinetic energy at the moment of the throw to potential energy at the moment of the highest point. If you raise the ball vertically upward at speed $v$, then $\dfrac{mv^2}{2} = mgh$, hence $h = \dfrac{v^2}{g}$, for example, if you throw the ball at speed $60 \frac{\text{mile}}{h} = 96.5 \frac{km}{h} = 26.8 \frac{\text{meter}}{s}$ then the ball will take off to a height of $\mathrm{73m = 76.5 \text{yards}}$

The acceleration due to gravity practically does not change near the earth and therefore it is considered constant. If the ball is thrown not high, then it will also accelerate 9.8 m/(s^2) (even astronauts on the ISS are affected by the acceleration of gravity of about 9.2 m/(s^2)). How to determine the height to which the ball will rise: equate the kinetic energy at the moment of the throw to potential energy at the moment of the highest point. If you raise the ball vertically upward at speed v, then m(v^2)/2 = mgh, hence h = (v^2)/g, for example, if you throw the ball at speed 60 mile/h = 96.5 km/h = 26.8 meter/s then the ball will take off to a height of 73 meter = 76.5 yards

The acceleration due to gravity practically does not change near the earth and therefore it is considered constant. If the ball is thrown not high, then it will also accelerate $9.8 \frac{m}{s^2}$ (even astronauts on the ISS are affected by the acceleration of gravity of about $9.2 \frac{m}{s^2}$). How to determine the height to which the ball will rise: equate the kinetic energy at the moment of the throw to potential energy at the moment of the highest point. If you raise the ball vertically upward at speed $v$, then $m*\frac{v^2}{2} = mgh$, hence $h = \frac{v^2}{g}$, for example, if you throw the ball at speed $60 \frac{mile}{h} = 96.5 \frac{km}{h} = 26.8 \frac{meter}{s}$ then the ball will take off to a height of $73 meter = 76.5 yards$