Suppose the ball is thrown at an angle of 33.0° below the horizontal. The point

Question

Suppose the ball is thrown at an angle of 33.0° below the horizontal. The point of release is h = 41.0 m above the ground.If it strikes the ground 39.0 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

in horizontal, a = 0

horizontal velocity = v cos33

t = 39 / v cos33

in vertical,

displacement, y = 0 - 41 = -41 m

initial vertical velocity. v0y = - v sin33

ay = - 9.8 m/s^2

in vertical,

-41 = - v sin33 t - g t^2 / 2

41 = v sin33 ( 39 / v cos33) + g (39 / v cos33)^2 / 2

41 = 25.3 + 10596/ v^2

v = 26 m/s

(A) t = 39 / (26 cos33) = 1.79 sec

(B) v = 26 m/s

(C) vfx = 26 cos33 = 21.8 m/s

vfy = v0y -g t = (- 26 sin33) - (9.8 x 1.79)

vfy = - 31.7 m/s

speed = sqrt(vfx^2 + vfy^2) = 38.5 m/s

angle = tan^-1(31.7 / 21.8) = 55 deg

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