Question

5a) A projectile is fired vertically upwards with height h given by: h = 25t - 4.9t^2 where 0 ≤ t ≤ 5

Find expressions for the velocity and acceleration. Determine the maximum

height reached by the projectile.

Answer 1

Given that $h = 25t - 4.9t^2$ and that $0 \leq t \leq 5$ , we have that

The velocity v, is $v = \dfrac{dh}{dt} = 25 - 9.8t$

Also, the acceleration a, is$a = \dfrac{dv}{dt} = \dfrac{d^2h}{dt^2} = -9.8$

The projectile will attain it maximum height when v = 0 i.e $v = \dfrac{dh}{dt} = 25-9.8t = 0\\ 9.8t = 25 \\ t = \frac{25}{9.8}$

Substituting into h(t), we have that

$h = 25(\frac{25}{9.8}) - 4.9(\frac{25}{9.8})^2\\ h = \frac{625}{9.8}(1- 0.5) \\ h = \frac{312.5}{9.8} = 31.8878$

Hence, the maximum height reached by the projectile is 31.8878m