Actually Welcome to the Concept of the Projectile Motion.
Since, here given that, vertical velocity= 50m/s
we know that u*sin(theta) = vertical velocity
so the time taken to reach the maximum height or the time of Ascent is equal to
T = Usin(theta) ÷ g, here g = 9.8 m/s^2
so we get as,
T = 50/9.8
T = 5.10 seconds
thus the time taken to reach max height is 5.10 seconds.
The position compared to that of home is a reference to displacement, I believe.
Displacement = x total - x initial
So I believe the answer is 5 blocks due north (if you’re walking linearly from your home), unless the questions is referring to relative displacement, in which then you’d need to use the Pythagorean theorem to find the hypotenuse between both positions. And then you’d have to find theta for the degrees between the south direction and the other unmentioned direction. But I don’t think that’s the case.
Distance refers to x total and doesn’t care for direction, as this refers to a scalar quantity opposed to a vector. Thus the equation is just
d = x
So 8 blocks + 3 blocks = a distance of eleven blocks walked total
The formula is=1/2(m x v^2)
so = 1/2*(0.05)*(310)^2
ans is =2402.5 joules
