Question

A rocket is launched straight up from the side of a cliff 43.89 m high. If the initial velocity is 34.14 m/sec and the height is given by 4the formula H=-16tsquared + vt + h. Find time the rocket will hit the ground.

Answer 1

Answer:

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Answer 2

Answer:

The time that the rocket will hit the ground is after $25.96 seconds.$

Step-by-step explanation:

You know that the height (H) is given by the formula $H= −16*t² + v*t+ h$

A rocket is launched directly from the side of a $43.89 m$ high cliff and that the initial velocity is $34.14 m / sec.$ So this indicates that the value of v is 34.14 m / sec and h is the height of $43.89 m$ . Replacing this value in the formula you get:

$H= −16*t² +34.14*t + 43.89$  

You must find the moment when the rocket will hit the ground, that is, the moment when the height H is zero:

$0= −16*t² +34.14*t + 43.89$

To calculate the time t the expression is used:

Being the general quadratic equation $ax² + bx + c = 0$, in this case $a=-16, b= 34.14 and c= 43.89.$  

Replacing:

Solving:

we get

$t1 = -1.2$

$t2 = 2.1$

Then:

or  

Solving for$t1$:

$t1=25.96$

Solving for $t2$:

$t2=-98.36$

Since the time cannot be negative, the time that the rocket will hit the ground is after $25.96 seconds.$