Lab 3 investigating projectile motion Need help on 7A and 7D LAB #3 INVESTIGATIN

Question

Lab 3 investigating projectile motion Need help on 7A and 7D LAB #3 INVESTIGATING PROJECTILE MOTION PHYS 1 110 CONCEPTUAL PHYSICS Instructor: Results: 9t) GETTING STARTED Name: Date: Open Google Chrome or other compatible browser and disable all browser pop-up blockers Go to https//phet.colorado.edu/en/simulation/legacy/prolectile motion Click Play symbol and select "OK" in the pop-up dialog box OBIECTIVES "Predict how varying conditions will effect the path of a projectile and explain your reasoning. Use the PhET simulation to confirm or disprove your predictions. Explain why your predictions were correct or incorrect. DATA COLLECTION 1. Look at the "height" box at the top. What kinematics term does it actually stand for, and what point is it relative to? (Hint: shoot the cannon once and watch the number closely). Explain your ng. 2. Fire the projectile launcher straight upwards (angle-90) at 18 m/s. Using kinematics, determine: a. the time it should take the projectile to reach maximum height gm1.2t -q.gm/s2 d-7 ? b. the maximum height reached by the projectile alaue cauron s leve Now, using the measuring tape, measure the actual height reached by the projectile (remember to measure from the little plus sign at the base of the cannon). Was your answer to (b) the same as this measurement? Ifit wasn't, check your math over and find your mistake. c. Tf the calauations d measuremens were done covecuy it hauld be the Same

Explanation / Answer

7a)

Let, ? be the angle of projection with horizontal line then we have horizontal component 18cos? & vertical component 18sin?.

Let the t be the time to hit the object then horizontal range

=(18cos?)(t)=R let R be the distance in x direction between cannon and target

?t=R/18cos?

In the same time tt projectile covers net 4.50-3.04=1.46m vetical height then we have

h=usin?+1/2gt2h

Substituting the corresponding values we get

1.46=?18sin?+1/2(9.81)(R/18cos?)2

from this equation we can calculate the required angle.

7d) QUESTION

Two balls A and B of masses 100g and 300g respectively are pushed horizontally from a table of height 3 meters. Ball has is pushed so that its initial velocity is 10 m/s and ball B is pushed so that its initial velocity is 15 m/s.

a) Find the time each ball takes to hit the ground.

b) What is the difference between the points of impact of the two balls on the ground?

ANSWER

The two balls are subject to the same gravitational acceleration and will hit the ground at the same time t ,

-3 = -(1/2) g t2

t = (3(2)/9.8) = 0.78 s

b) Horizontal distance XA of ball A

XA = 10 m/s * 0.78 s = 7.8 m

Horizontal distance XB of ball B

XB = 15 m/s * 0.78 s = 11.7 m

Difference in distance XA and XB is given by

|XB - XA| = |11.7 - 7.8| = 3.9 m

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