Question

A rock is thrown upward with a velocity of 20 meters per second from the top of a 24 meter high cliff, and it misses the cliff on the way back down. When will the rock be 10 meters from the water, below? Round your answer to two decimal places

use the gravity formula

Its a cluster in my brain

Answer 1

Answer:

(a)

$h(t)=-4.9t^2+20t+24$

(b)

$t=4.69sec$

Step-by-step explanation:

(a)

We are given

velocity is 20 meters per second

$v_0=20m/s$

the top of a 24 meter high cliff

so,

$h_0=24$

now, we can use gravity formula

$h(t)=-4.9t^2+v_0t+h_0$

now, we can plug value

$h(t)=-4.9t^2+20t+24$

(b)

we are given

h(t)=10

we can set it and then we can solve for t

$h(t)=-4.9t^2+20t+24=10$

$-4.9t^2\cdot \:10+20t\cdot \:10+24\cdot \:10=10\cdot \:10$

$-49t^2+200t+140=0$

now, we can use quadratic formula

$ax^2+bx+c=0$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

we can compare and find a,b and c

a=-49 , b=200 , c=100

$t=\frac{-200\pm \sqrt{200^2-4\left(-49\right)140}}{2\left(-49\right)}$

$t=-\frac{2\left(\sqrt{4215}-50\right)}{49},\:t=\frac{2\left(50+\sqrt{4215}\right)}{49}$

we get

$t=-0.609,t=4.691$

we know that

time can never be negative

so,

$t=4.691sec$

Correct to 2 decimal places

$t=4.69sec$