3))) The position of a particle as a function of time is given by 6.60 2.30 , wh

Question

3))) The position of a particle as a function of time is given by 6.60 2.30 , where is in seconds. What is the particle's distance from the origin at s? What is the particle's distance from the origin at = 3.00 ? What is the particle's distance from the origin at = 5.90 ? 5) A ball thrown horizontally at 27 travels a horizontal distance of 57 before hitting the ground. From what height was the ball thrown? 6)) A supply plane needs to drop a package of food to scientists working on a glacier in Greenland. The plane flies 90.0 above the glacier at a speed of 120 . How far short of the target should it drop the package? 7)) In the Olympic shotput event, an athlete throws the shot with an initial speed of 12 at a 40.0 angle from the horizontal. The shot leaves her hand at a height of 1.8 above the ground. How far does the shot travel? Repeat the calculation of part (a) for angles of 42.5 , 45.0 , and 47.5 . Express your answer to four significant figures and include the appropriate units. Express your answer to four significant figures and include the appropriate units.

Explanation / Answer

5)57m / 27m/s = 2 .1sec

2.1 = sqrt(2h/9.8)

4.41 = 2h/9.8

43.218 = 2h

h = 21.609 m

6)d = 1/2(a)(t^2)
900 = 1/2(9.8)(t^2)
t = 13.55s
It takes 13.55 seconds for the package to hit the ground.
13.55s before the scientists is (13.55s)(120m/s) = 1626m

7)X0 = 0m
T0 = 0s
Y0 = 1.8m
V0 = 12m/s
Ay = -9.8m/s^2
Angle = 40degree

V0x = V0cos(40)
V0y = V0sin(40)

Y1 = Y0 + V0y*(T1-T0) - 0.5g(T1)^2
0 = 1.8 + 12sin(40)*(T1) - 0.5(9.8)(T1)^2
0 = 1.8 + 7.71(T1) - 4.9(T1)^2
Solve the quadratic equation, you'll get 1.78s for T1

X = (V0x)(T1) = 12cos40(1.78) = 16.36m

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