Answer:
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
. The smaller leg is the hypotenuse of a right triangle sides 4 and 
so the usual trig cliche, hypotenuse 8.