Suppose the ball is thrown from the same height as in the PRACTICE IT(43.0m) pro

Question

Suppose the ball is thrown from the same height as in the PRACTICE IT(43.0m) problem at an angle of 31.0°below the horizontal. If it strikes the ground 44.6 m away, find the following. (Hint: For part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.)

(a) the time of flight_______s

(b) the initial speed______ m/s

(c) the speed and angle of the velocity vector with respect to the horizontal at impact

speed ______ m/s

angle ________° below the horizontal

Explanation / Answer

along horizantal

initial velocity vox = vo*cos31

ax = 0

x = vox*T

X = vo*cos31*T

vo*T = X/cos31 .....(1)

along vertical

voy = -vo*sin31

ay = -g

y = voy*T + 0.5*ay*T^2

y = -vo*T*sin31- 0.5*g*T^2 ..........(2)

y = -43

x = 44.6

y = -xtan31 - 0.5*9.8*T^2

-43 = -44.6*tan31*T - 0.5*9.8*T^2

T = 1.3 s

+++++

(b)

vo = X/T*cos31

vo = 44.6/(1.3*cos31)

vo = 40

+++++++++++

(c)

vy = voy + ay*T

vy = -40*sin31*1.3 - 9.8*1.3 = -39.5 m/s

vx = vox + ax*T = vox = 40*cos31 = 34.3

speed = sqrt(39.5^2+34.3^2) = 52.3

direction = tan^-1(39.5/34.3) = 49

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