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

answer: the answer is 30 miles per hour.

ft/s -> mphmph = ft/s 1.466667mph = 44 divided by 1.466667mph = 30.00000

- r3ktforgood ☕