Answer:
26.5 units
Step-by-step explanation:
The height of the ball at any time t is given by :
[tex]h(t)= -16t^2 + 40t + 1.5[/tex] ...(1)
We need to find the maximum height of the ball.
For maximum height, put [tex]\dfrac{dh}{dt}=0[/tex]
i.e.
[tex]\dfrac{d( -16t^2 + 40t + 1.5)}{dt}=0\\\\-32t+40=0\\\\t=\dfrac{40}{32}\\\\t=1.25\ s[/tex]
Put t = 1.25 in equation (1) to find the maximum height.
So,
[tex]h(t)= -16(1.25)^2 + 40(1.25) + 1.5\\\\=26.5\ units[/tex]
So, the maximum height reached by the ball is 26.5 units.