a player kicks ball into air with an initial velocity of 30m/s at an angle of 60° find the range of the ball
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The ball’s initial velocity of 10 m/s can be broken down into a horizontal component vx0vx0, and a vertical component, vy0vy0.
vx0=(10cos30°)vx0=(10cos30°)m/s, or about 8.66 m/s
vy0=(10sin30°)vy0=(10sin30°) m/s, or 5 m/s
Neglecting air resistance — which we will definitely be there, as addressing it would be well outside the scope of this simple problem — vxvx is constant, and vyvy changes due to gravitational acceleration:
vx(t)=vx0=(10cos30°)vx(t)=vx0=(10cos30°) m/s
vy(t)=vy0+gtvy(t)=vy0+gt, where g=−9.81g=−9.81 m/s22— negative since +y is positive, thus (5−9.81t)(5−9.81t) m/s
At t=1 s, vx=(10cos30°)vx=(10cos30°) m/s ≈≈ 8.66 m/s, while vy=(5−9.81(1))vy=(5−9.81(1)) m/s =−4.81=−4.81 m/s
Combining those components, v=v2x+v2y−−
vx0=(10cos30°)vx0=(10cos30°)m/s, or about 8.66 m/s
vy0=(10sin30°)vy0=(10sin30°) m/s, or 5 m/s
Neglecting air resistance — which we will definitely be there, as addressing it would be well outside the scope of this simple problem — vxvx is constant, and vyvy changes due to gravitational acceleration:
vx(t)=vx0=(10cos30°)vx(t)=vx0=(10cos30°) m/s
vy(t)=vy0+gtvy(t)=vy0+gt, where g=−9.81g=−9.81 m/s22— negative since +y is positive, thus (5−9.81t)(5−9.81t) m/s
At t=1 s, vx=(10cos30°)vx=(10cos30°) m/s ≈≈ 8.66 m/s, while vy=(5−9.81(1))vy=(5−9.81(1)) m/s =−4.81=−4.81 m/s
Combining those components, v=v2x+v2y−−
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