Let's first find x and y components of the initial velocity of the ball:
v0x=v0cosθ=8 sm⋅cos30∘=6.93 sm,v0y=v0sinθ=8 sm⋅sin30∘=4 sm.Then, we can find the time that the ball takes to reach the ground from the kinematic equation. Let's take the upwards as the positive direction, then we get:
y=y0+v0yt−21gt2,0=10−4t−4.9t2,4.9t2+4t−10=0.This quadratic equation has two roots: t1=1.08 and t2=−1.89. Since time can't be negative the correct answer is t=1.08 s.
Finally, we can find the range of the ball:
x=v0xt=6.93 sm⋅1.08 s=7.48 m.
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