At time t1 and t2, a projectile projected vertically up reaches the same height H from the ground. Find the ma

At time t1 and t2, a projectile projected vertically up reaches the same height H from the ground.
Find the maximum height and time to reach the ground.

let t2 >t1.

let t2 >t1.

In (t2- t1) /2 it will reach the maximum height.
The distance traveled in this time is g(t2- t1) ²/8

The total height is H + g(t2- t1) ²/8
=========================

Time of flight is the time to reach the ground = t1 + t2 + t1 = 2t1 + t2.
=============================

We can write the equation describing the object's height as a function of time directly from the given data:

h(t) = H - (g/2)( t - t1 )( t - t2 )

Since this is an inverted parabola that passes through (t1,H) and (t2,H) and has a quadratic coefficient of -(g/2). By symmetry, the maximum height occurs at the midpoint of the two given times:

h( (t1+t2)/2 ) = H - (g/2) ( t2 - t1 )/2 ( t1 - t2 )/2 = H + (g/8) (t2 - t1)^2 <== First answer

The object will hit the ground when h(t)=0:

0 = 2H/g - ( t^2 - (t1 + t2)t + t1t2 )

0 = t^2 - (t1 + t2)t + ( t1t2 - 2H/g )

Which we can solve using quadratic formula:

t = [ (t1 + t2) +/- sqrt( (t1 + t2)^2 - 4(t1t2 - 2H/g) ) ] / 2

t = (t1 + t2)/2 +/- sqrt( (t2 - t1)^2 + 8H/g )/2

Which makes sense because it is two points in time that are symmetric about the midpoint of the equiheight times. Thus the total time of flight is double the +/- term:

sqrt( (t2 - t1)^2 + 8H/g ) <<= Second answer