Projectile Motion - Physics Question?

Question

A new circus act is called the Texas Tumblers. Lovely Alex swings from a trapeze, projects herself at an angle of 53 degrees, and is supposed to be caught by Dylan, whose hands are 6.1 m above and 8.2 meters horizontally from Alex's launch point.

a. What initial speed must Alex have to reach Dylan?

b. For the initial speed calculated in Part a, what are the magnitude and direction of Alex's velocity when she reaches Dylan?

c. The night of their performance, Dylan misses her completely as she flies past. How far horizontally does Alex travel, from her initial launch point before landing in the safety net 8.6 m below her starting point?

Answer 1

x = 8.2

y = 6.1

a = 53 degrees

Vh = Vcos 53 = 8.2/t

t = 8.2/(Vcos 53)

Vy = Vsin 53

y = Vy t - 0.5gt^2

6.1 = Vsin 53 (8.2/V cos 53) - 0.5(9.8)(8.2/V cos 53)^2

6.1 = tan 53 (8.2) - 0.5(9.8)*(8.2)^2/(V^2 cos^2 53)

0.5(9.8)*(8.2)^2/(V^2 cos^2 53)=  8.2 tan 53-  6.1

329.5/ 0.362V^2 = 10.9 -6.1

329.5/ 0.362V^2 = 4.8

0.362V^2 = 329.5/ 4.8

0.362V^2 = 68.6

V^2 = 68.6/0.362

V^2 = 189.6

V = 13.77 m/s

b)

t = 8.2/(Vcos 53)

t = 8.2/(13.77cos 53)

t = 8.2/8.28

t = 1.0 sec

At t = 1 sec

Vh = Vcos53 (in projectile motion the horiz component of V is constant)

Vh = 13.77 cos 53 = 8.28 m/s

Vy = Vo - gt

Vy = 13.77 sin 53 - 9.8(1)

Vy = 11.0 - 9.8

Vy = 1.2 m/s

the resultant velocity at 1 sec

V^2 = (Vx^2 + Vy^2) = 8.28^2 + 1.2^2

V^2 = 67.6 + 1.44

V = sqrt(69)

V = 8.3 m/s

direction

tan Vy/Vx = tan (1.2/8.28)

tan a = 0.129

a = 7.3 degrees

c.

at maximum height Vy = 0, Vx = 8.28 (constant)

intial Vy= Voy = Vsin 53 = 13.77*(0.799)

Voy = 10.9 m/s

Vf^2 - Voy^2 = -2gy

0 - 13.77^2 = -2(9.8)y

y = -13.77^2/ -2(9.8)

maximum height

y = -189.6/-19.6

y = 9.7 m

from this information of  max height, we can calculate the time for her to fall from peak of flight down to launch point (initial vertical position) then to a point 8.6 m below the initial.

the time to fall back to initial point is equal to the time going up to the maximum height (right?) time to 9.7 m up and 9.7 m going down to position of launch.

If we take the downward path, Voy = 0

ymax = Vot - 1/2 gt^2

-9.7 = 0 - 1/2 (9.8) t^2

-9.7 = -4.9 t^2

t^2 = -9.7/-4.9

t = sqrt(1.98)

t =1.4 sec

t = 1.4 sec is also the time from launch pad to peak of flight, lets call this t1

t1 = 1.4 sec

now compute the time of fall from peak of flight down to 8.6 m below lauch position.

from peak to initial launch position = 9.7 m

but she fell 8.6 m below this point where the net is

total height of fall from peak is 9.7 + 8.6 = 18.3 m

-18.3 = Voy t - 1/2 gt^2

-18.3 = 0 - 4.9 t^2

t^2 = -18.3/-4.9

t = sqrt(3.75)

t= 1.9 sec

lets call this t2

total time t = t1 + t2 = 1.4 + 1.9 = 3.3 sec

t1 is the upward path and t2 is the downward path

horizontal distance from launch point (remember in the x direction V is constant = 8.28 m/s)

x = Vox t

x = 8.28(3.3)

x = 27.3 m

poor Alex, hope the net is strong enough to stop her momentum