106k views
1 vote
PLEASE HELP WITH THESE TWO QUESTIONS- SHOW WORK ! THANK YOU !

For a projectile fired at 28 degrees from a height of 0 m, how long does it take to hit the ground? Remember that a_y=-9.8 m/s^2. Initial velocity of the projectile is 42 m/s. And second question is - using your answer from that above question, how far did the cannonball travel in the horizontal direction?

1 Answer

4 votes

Answer:

4.06 seconds

148.3 meters

Step-by-step explanation:

To solve for the time it takes for the projectile to hit the ground, we can use the following kinematic equation:

y = y_0 + v_0y * t + 1/2 * a_y * t^2

where y is the final height (0 m), y_0 is the initial height (also 0 m), v_0y is the initial vertical velocity, and a_y is the acceleration due to gravity (-9.8 m/s^2).

First, we need to find v_0y, the initial vertical velocity:

v_0y = v_0 * sin(theta) = 42 m/s * sin(28 degrees) ≈ 19.6 m/s

Now, we can solve for the time it takes to hit the ground:

0 = 0 + 19.6 m/s * t + 1/2 * (-9.8 m/s^2) * t^2

Simplifying and solving for t, we get:

t ≈ 4.06 seconds

Therefore, it takes about 4.06 seconds for the projectile to hit the ground.

To find the horizontal distance traveled by the projectile, we can use the following kinematic equation:

x = v_0x * t

where x is the horizontal distance traveled, v_0x is the initial horizontal velocity, and t is the time it takes to hit the ground (which we just calculated).

Since there is no air resistance, the horizontal velocity remains constant throughout the motion. Therefore:

v_0x = v_0 * cos(theta) = 42 m/s * cos(28 degrees) ≈ 36.5 m/s

Now, we can plug in the values to find the horizontal distance traveled:

x = 36.5 m/s * 4.06 s ≈ 148.3 meters