Agrasshopper jumps straight up from the ground with an initial vertical velocity of 8 feet per second. after how many seconds will the grasshopper be 1 foot off the ground? a squirrel is 27 feet up in a tree and tosses a nut out of the tree with an initial vertical velocity of 6 feet per second. the squirrel climbs down the tree in 2 seconds. does it reach the ground before the nut? i don’t get the process of solving these types of problems me?

Problem 1 : After 0.25 s

Problem 2 : The squirrel doesn't reach the ground before the nut

Step-by-step explanation:

Problem 1

This is a standard physics problem

The equation that defines the movement of the grasshopper given the initial velocity, and the acceleration is the following

Distance =  Do + Vo*t + 0.5*a*t^2

Where

Do is the initial distance.

(Do = 0, because the grasshopper is on the ground and we choose the reference system there)

Vo is the initial velocity = 8 feet/sec

a is the acceleration that is exerted on the body (in this case a = g = -9.8 m/s^2 = - 32.174 feet per second per second.)

t is the time

We substitute in the equation, and solve for t (time)

Distance =  Do + Vo*t + 0.5*a*t^2

(1 foot)=  (0)+ (8 feet/s)*t + 0.5*(-32.174 feet/s^2)*t^2

Solving this equation, we get that

t = 0.25 s

Problem 2

This problem is similar to the previous one

We apply the same formula

D =  Do + Vo*t + 0.5*a*t^2

But in this case, we are going in the opposite direction, and we have an initial distance.

Do = 27 feet

Vo is the initial velocity of the nut = -6 feet/s

D = 0  (Ground)

We substitute in the equation, and solve for t (time)

(0) =  (27) + (-6 feet/s)*t + 0.5*(-32 feet/s^2)*t^2

t = 1.125 s  <  2s

The nut reaches ground in 1.125 s, so the squirrel doesn't reach the ground before the nut

question

what is the value of the third quartile of the data set represented by this box plot? a. 19 b. 21 c. 26 d. 29

answer

c

false

step-by-step explanation: