Use the worked example above to help you solve this problem. A ball is thrown up

Question

Use the worked example above to help you solve this problem. A ball is thrown upward from the top of a building at an angle of 30.0degrees to the horizontal and with an initial speed of 18.0 m/s. The point of release is h = 47.0 m above the ground. How long does it take for the ball to hit the ground? 4.15 s Find the ball's speed at impact. Enter a number m/s Find the horizontal range of the ball. m GETTING STARTED | I'M STUCK! Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 25.0degrees below the horizontal. If it strikes the ground 60.4 m away, find the following. the time of flight s the initial speed m/s the speed and angle of the velocity vector with respect to the horizontal at impact speed m/s angle degrees below the horizontal

Explanation / Answer

60.4 = v0 * cos 250 * t

v0 * t = 66.64

h = -v0 sin250* t - 0.5gt2

h = -66.64*0.423 - 0.5*9.8*t2 h = - 47

(-47+28.19)/(-0.5*9.8) = t2

t = 1.96 sec

v0t = 66.64 , therefore v0 = 34 m/s

Horizontal speed will remail same as v0 cos 25 = 30.81 m/s

Vertical speed = -v0 sin 25 - gt = -33.58 m/s

Velocity when it hits ground = 30.81 i - 33.58 j

Speed = (30.812 + 33.582) 0.5 = 45.57 m/s

Angle = tan-1 (33.58/30.81) = 47.460

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