Question

A rocket is launched from a height of 3 m with an
initial velocity of 15 m/s.
What is the maximum height of the rocket? When
will this occur?

Answer 1

Answer:

14.25 m

1.5 s

Explanation:

Given:

y₀ = 3 m

v₀ = 15 m/s

v = 0 m/s

a = -10 m/s²

Find: y and t

v² = v₀² + 2aΔy

(0 m/s)² = (15 m/s)² + 2 (-10 m/s²) (y − 3 m)

y = 14.25 m

v = at + v₀

0 m/s = (-10 m/s²) t + 15 m/s

t = 1.5 s

Answer 2

The maximum height reached by rocket is 14.48m

The time the rocket will reach its maximum height is 1.53 secs

To get the maximum height, we will use the equation of motion. Using the equation of motion;

$v^2=u^2-2g(\triangle s)$

where:

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

$\triangle s$ is the change in height

The equation can also be written as $v^2=u^2-2g(s_2-s_1)$

Substituting the given parameters into the formula:

$0^2=15^2-2(9.8)(s_2-3)\\0 = 225 - 19.6s_2+58.8\\19.6s_2=283.8\\s_2=\frac{283.8}{19.6}\\s_2= 14.48m$

Hence the maximum height reached by rocket is 14.48m

Get the required time this will occur.

$v=u-gt\\0=15-9.8t\\9.8t=15\\t=\frac{15}{9.8}\\t= 1.53secs$

Hence the time the rocket will reach its maximum height is 1.53 secs

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