A diver dives into the sea from a height of 35m. His height h meters t seconds after leaving the cliff is given by h = -4.9² + t + 35. How long until he reaches the water?

Jul 28, 2016

2.67 seconds

Explanation:

The diver starts at the height of 35 metres. At the point of impact with the water his height is 0.

$0 = - {4.9}^{2} + t + 35$

color(red)("Is this equation correct?"

As I understand things

Acceleration will be 9.81 metres per second squared

So height will be starting distance -( mean velocity times time)

mean velocity $\to \frac{1}{2} a t \to \text{distance} = \frac{1}{2} \left(9.81\right) {t}^{2}$

So height $\to h = 35 - \frac{1}{2} \left(9.81\right) {t}^{2}$

$\implies 0 = 35 - \frac{1}{2} \left(9.81\right) {t}^{2}$

$\frac{1}{2} \left(9.81\right) {t}^{2} = 35$

${t}^{2} = \frac{70}{9.81}$

$\implies t = \sqrt{\frac{70}{9.81}} \approx 2.67 \text{ seconds}$
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Standard equation is $s = u t + \frac{1}{2} a {t}^{2}$

$u = 0$

$s = \frac{1}{2} a {t}^{2} = 4.9 {t}^{2} \text{ " larr 1/2a = 9.8/2=4.905 " call it } 4.9$

so $h = 35 - 4.9 {t}^{2}$