Acannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. the shell’s trajectory curves downward because of gravity, so at the time t = 0.555 s the shell is below the straight line by some vertical distance ∆h. find this distance ∆h in the absence of air resistance. the acceleration of gravity is 9.8 m/s 2 .

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

Vy in this problem will follow this equation =

where g is the gravity acceleration

This is equation (1)

For Y(t) :

We suppose yi = 0

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

So in t = 0.675 s  → Vy = 0. Now we calculate the y in which this happen using (2)

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

answer: (431,2)

step-by-step explanation:

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be