Answer:
a) First let's analyze the vertical problem:
When the projectile is on the air, the only vertical force acting on it is the gravitational force, then the acceleration of the projectile is the gravitational acceleration, and we can write this as:
a(t) = -9.8m/s^2
To get the vertical velocity we need to integrate over time to get:
v(t) = (-9.8m/s^2)*t + v0
where v0 is the initial vertical velocity because the object is thrown horizontally, we do not have any initial vertical velocity, then v0 = 0m/s
v(t) = (-9.8m/s^2)*t
To get the vertical position equation we need to integrate over time again, to get:
p(t) = (1/2)*(-9.8m/s^2)*t^2 + p0
where p0 is the initial position, in this case is the height of the edifice, 20m
then:
p(t) = (-4.9m/s^2)*t^2+ 20m
The projectile will hit the ground when p(t) = 0m, then we need to solve:
(-4.9m/s^2)*t^2+ 20m = 0m
20m = (4.9m/s^2)*t^2
√(20m/ (4.9m/s^2)) = t = 2.02 seconds
The correct option is a.
b) The range will be the total horizontal distance traveled by the projectile, as we do not have any horizontal force, we know that the horizontal velocity is 25 m/s constant.
Now we can use the relationship:
distance = speed*time
We know that the projectile travels for 2.02 seconds, then the total distance that it travels is:
distance = 2.02s*25m/s = 50.5m
Here the correct option is a.
c) Again, the horizontal velocity never changes, is 25m/s constantly, then here the correct option is option b. 25m/s
d) Here we need to evaluate the velocity equation in t = 2.02 seconds, this is the velocity of the projectile when it hits the ground.
v(2.02s) = (-9.8m/s^2)*2.02s = -19.796 m/s
The velocity is negative because it goes down, and it matches with option d, so I suppose that the correct option here is option d (because the sign depends on how you think the problem)
