Question

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?

Answer 1

0.969 s (3 s.f.)

Step-by-step explanation:

Constant Acceleration Equations (SUVAT)

`\boxed{\begin{array}{c}\begin{aligned}v&=u+at\\\\s&=ut+(1)/(2)at^2\\\\ s&=\left((u+v)/(2)\right)t\\\\v^2&=u^2+2as\\\\s&=vt-(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}}`

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.

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

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