Question

with that speed in mi/hr must an object be thrown to reach a height of 91.5m(equivalent to one football field?) assume negligible air resistance and 1m/s+2.23 mi/hr.

Answer 1

\[\Delta x = Vi \Delta t + 1/2a \Delta t ^{2}\]

Answer 2

You could use an energy conservation argument, i e that kinetic energy at ground level (where potential energy is zero) is equal to potential energy at 91.5 m (standstill at max height - kinetic energy is zero). \[mv ^{2}/2=mgh => v ^{2}/2=gh => v=\sqrt{2gh}\]

Answer 3

v^2=u^2 +2as => 0=u^2 +2*-9.8*91.5 => u^2=1793.4
therefore u= root (1793.4) = 42.3 ms^-1 = (i think 1 km/hr = 1.61mile/hr) 94.6 mile/hr


that last figure moight be wrong but the speed in m/sec is correct.