7.00 110 5.00 WCD Physies 235 Exam#1 NAME : 13. A stunt woman is launched with a

Question

7.00 110 5.00 WCD Physies 235 Exam#1 NAME : 13. A stunt woman is launched with an initial speed of 20 ms at an angle of 53 above the horizontal from the top of a 55 m tall building. If g = 9.8 m sz and air resistance is negligible, then what is the magnitude ofthe horizontal component of velocity as she lands on the safety cushion at ground level. a) 7.5 m/s b) 9.0 m/s c) 12 m's d) 29 m/s v 20 m's 53° 55 m 14. What was the magnitude of the vertical component of velocity as sbe strikes the cushion? a) 25 m's b) 36 ms c) 49 m's d) 144 m's 15. How long after being launched does she reach ber maximum altitude? a) 1.5s b) 1.6s .7s) 1.8s 16. What maximum altitude does the she reach? a) 60 m b) 68 m )77 md) 120 m 3 of 4

Explanation / Answer

Solution :-

(1)

horizontal velocity remains constant through out the projectilr motion, since there is no horizontal acceleration

So,

Vxf = Vxi = 20*cos 53 deg = 12 m/sec

(2)

Since at max height Vy = 0

ay = + 9.81 m/sec^2

Hmax = 68 m from 4th part

using the equation

Vyf^2 = Vy^2 + 2*a*Hmax

Vyf = sqrt (0^2 + 2*9.81*68) = 36.52 m/sec

(3)

Time to reach the max height = t1 = V0*sin theta/g

t1 = 20*sin 53 deg/9.81 = 1.6 sec

(4)

Hmax = V0^2*(sin theta)^2/()2*g + 15 + 40

Hmax = 20^2*(sin 53 deg)^2/(2*9.81) + 15 + 40 = 68 m

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