Answer:
the cannonball’s velocity parallel to the ground is 86.6m/S
Explanation:
Hello! To solve this problem remember that in a parabolic movement the horizontal component X of the velocity of the cannonball is constant while the vertical one varies with constant acceleration.
For this case we must draw the velocity triangle and find the component in X(see atached image).
V= Initial velocity=100M/S

V= Initial velocity=100M/S
Vx=cannonball’s velocity parallel to the ground
Solving for Vx
Vx=Vcos30
Vx=(100m/S)(cos30)=86.6m/s
the cannonball’s velocity parallel to the ground is 86.6m/S
1 inch = 2.54cm. 11*2.54 = 27.94cm. You would use multiplication. Hope this helps!! Can I get brainliest pls
Answer:
A. Area under force-time graph & Area under force-displacement graph
Explanation:
To find the impulse or work done the area under force-time graph and area under force-displacement graph will give us these respective values.
Impulse = Force x time
Work done = Force x displacement
When we plot a graph of force and time, the area under it is the impulse.
When a graph of force and displacement is plotted, the area under is the work done.
<h2>Hope it's helpful for you ✌️✌️✌️✌️✌️</h2>