A tennis ball is thrown straight up into the air with an initial velocity of 9.5 m/s. How long does it take to reach the top of the trajectory?

[tex]\boxed{\begin{array}{c}\begin{aligned}v&=u+at\\\\s&=ut+\dfrac{1}{2}at^2\\\\ s&=\left(\dfrac{u+v}{2}\right)t\\\\v^2&=u^2+2as\\\\s&=vt-\dfrac{1}{2}at^2\end{aligned}\end{array}} \quad \boxed{\begin{minipage}{4.6 cm}$s$ = displacement in m\\\\$u$ = initial velocity in ms$^{-1}$\\\\$v$ = final velocity in ms$^{-1}$\\\\$a$ = acceleration in ms$^{-2}$\\\\$t$ = time in s (seconds)\end{minipage}}[/tex]

When using SUVAT, assume the object is modeled as a particle and that acceleration is constant.

Acceleration due to gravity: g = 9.8 ms⁻²

Resolving vertically, taking up as positive:

u = 9.5 m/s

v = 0 m/s
When projected objects reach the top of their motion, they stop momentarily. Therefore, the velocity v is zero at that point.

a = -9.8 m/s²
As g always acts downwards and up is taken as positive, acceleration is negative.

[tex]\begin{aligned}\textsf{Using} \quad v&=u+at:\\\\0&=9.5+(-9.8)t\\-9.5&=-9.8t\\t&=\dfrac{-9.5}{-9.8}\\t&=0.9693877551\\\implies t&=0.969\; \sf s\;(3\:s.f.)\end{aligned}[/tex]

Therefore, it takes 0.969 s (3 s.f.) for the tennis ball to teach the top of its trajectory.