Lab 4: Projectile Motion 3. Take the data from your best trial run and put the b

Question

Lab 4: Projectile Motion 3. Take the data from your best trial run and put the below corresponding values into the table Release Time in height | air… Distance | Vo | Voy | IVol | Horizontal Now, as in Lab 3, using your best trial data, write out the equations (also known as functions) that predict the horizontal position x(t) as a function of time, the vertical position as a function of time y(0), the horizontal velocity as a function of time vs(t), and the vertical velocity as a function of time vs(t)This part is the same as in Lab 3, except now, record functions to include horizontal motion along with the vertical motion. Remember, record functions for the data you actually took. These functions should include your value of Voy Yo and also Vs. The independent variable of the functions for both directions is the time "" For example, if asked, "What is the vertical coordinate of your ball y(t) and what is its vertical velocity v,(1) at 1.0 second?", the y(t) and v,(t) for your data below would compute a vertical coordinate and a vertical velocity to go with 1.0 second. Likewise, since you will now have two additional functions to quantify the horizontal direction, one of your horizontal functions will compute the horizontal coordinate, x(t) and the other horizontal function will compute a horizontal velocity v(t) at 1.0 seconds. (Note: it's okay if some functions are constant!) If asked, "What is the horizontal and vertical coordinates of your ball and what is the ball's horizontal and vertical velocity at 2.0 seconds?", your four functions below will compute both a horizontal and vertical velocity along with a vertical coordinate and horizontal coordinate to go with 2.0 seconds. Horizontal position: x(t)=Vot Horizontal velocity: v(t)VOX Vertical position: yo Vone Vertical velocity: v,(t)Vayt

Explanation / Answer

As per the data given by you:

h0 = 5.36 m ; t = 1.35 s ; D = 4.9 m ; v0x = 3.63 m/s

v0y = 9.3 m/s ; v0 = 9.98 m/s ; theta = 68.68 deg

The maximum height attained by the projectile is given by:

H = v0y^2/2g = 9.3^2/2 x 9.8 = 4.41 m

Hence, H = 4.41 m

We know that the horizontal velocity remainsa same through out the course of motion for a projectile.

But the vertical velocity changes

from eqn of motion

v = u + at

at t = 1 s (as in the paragraph, it ask to find vy at t = 1 s)

vy = 9.3 - 9.8 x 1 = -0.5 m/s

V = sqrt (v0x^2 + Vy^2) = sqrt(3.63^2 + 0.5^2) = 3.66 m/s

Hence, V = 3.66 m/s

a = acceleration remains same through out the course of motion.

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