Question

A body is projected upward at an angle of 30 degree to the horizontal at an initial speed of 200ms-.In how many seconds will it reach the ground? How far from the point of projection would it strike?

Answer 1

Answer:

20.41 s

3534.80 m

Explanation:

In how many seconds will it reach the ground?

We are given the initial velocity of the body, which is 200 m/s at a 30° angle.

We know the acceleration in the vertical direction is -9.8 m/s², assuming that the upwards/right direction is positive and the downwards/left direction is negative.

Since we are using acceleration in the y-direction, let's use the vertical component of the initial velocity.

• 200 · sin(30) m/s

Let's use the fact that at the top of its trajectory, the body will have a final velocity of 0 m/s.

Now we have one missing variable that we are trying to solve for: time t.

Find the constant acceleration equation that contains v₀, v, a, and t.

• v = v₀ + at

Substitute known values into the equation.

• 0 = 200 · sin(30) + (-9.8)t
• -200 · sin(30) = -9.8t
• t = 10.20408163

Recall that this is only half of the body's trajectory, so we need to double the time value we found to find the total time the body is in the air.

• 2t = 20.40816327

The body will reach the ground in 20.41 seconds.

How far from the point of projection would it strike?

We want to find the displacement in the x-direction for the body.

Let's find the constant acceleration equation that contains time t, that we just found, and displacement (Δx).

• Δx = v₀t + 1/2at²

Substitute known values into the equation. Remember that we want to use the horizontal component of the initial velocity and that the acceleration in the x-direction is 0 m/s².

• Δx = (200 · cos(30) · 20.40816327) + 1/2(0)(20.40816327)²
• Δx = 3534.797567

The body will strike 3534.80 m from the point of projection.