Answer:
<h3>1.03684m</h3>
Explanation:
Using the formula for calculating range expressed as;
R = U√2H/g where
R is the distance moves in horizontal direction = 18.4m
H is the height
U is the velocity of the baseball = 40m/s
g is the acceleration due to gravity = 9.8m/s²
Substitute the given parameters into the formula and calculate H as shown;
18.4 = 40√2H/9.8
18.4/40 = √2H/9.8
0.46 = √2H/9.8
square both sides;
(0.46)² = (√2H/9.8)²
0.2116 = 2H/9.8
2H = 9.8*0.2116
2H = 2.07368
H = 2.07368/2
H = 1.03684m
Hence the ball is 1.03684m below the launch height when it reached home plate.
Answer:
649kg/m^3
Explanation:
Let p be the density of this particular object.
Formula for density:
We can substitute the givenmass and volume to find density of the object.
Therefore the density of this object is 649kg/m^3.
Answer:
Velocity
Explanation:
"The principle is that the slope of the line on a position-time graph is equal to the velocity of the object. If the object is moving with a velocity of +4 m/s, then the slope of the line will be +4 m/s."
^^This explanation is from physicsclassroom.com
<span>25,000 miles per hour
hope that this helps</span>
Answer:
c. initial (x and y)
Explanation:
When a projectile is launched at a velocity with a launch angle, to solve it, we must first resolve the initial velocity into the x and y components. To do this will mean we have to treat it like a triangle due to the launch angle and the direction of the projectile.
Therefore, we will have to make use of trigonometric ratios which is also known by the mnemonic "SOH CAH TOA"
Thus, this method resolves the initial x and y velocities.
