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Projectile time of flight and range

Short answer

Resolve initial velocity along 3:4 slope: v_x = 24 m/s, v_y = 18 m/s. Vertical equation −25 = 18t − 4.905t² gives t = 4.74 s. Range = 24×4.74 = 113.9 m.

A projectile is projected as shown in figure with an initial velocity of 30 m/s. Determine the time of flight and range. [Figure: launch point A on a slope of 3 vertical to 4 horizontal (angle θ\theta) at the launch end; the projectile lands at ground point B, 25 m below A.]

Assumption (figure not shown): the slope at A is 3:43:4, so tan⁡θ=3/4\tan\theta=3/4, cos⁡θ=0.8\cos\theta=0.8, sin⁡θ=0.6\sin\theta=0.6. The 30 m/s launch is along the slope, upward. B is 25 m below A. Take g=9.81 m/s2g=9.81\ \text{m/s}^2 and neglect air resistance.

Components of launch velocity

vx=30(0.8)=24 m/s,vy=30(0.6)=18 m/sv_x=30(0.8)=24\ \text{m/s},\qquad v_y=30(0.6)=18\ \text{m/s}

Time of flight

Vertical motion from A to B (y=−25y=-25 m):

−25=18t−12(9.81)t24.905t2−18t−25=0t=18+182+4(4.905)(25)2(4.905)=4.744 s\begin{aligned} -25&=18t-\tfrac12(9.81)t^{2}\\ 4.905t^{2}-18t-25&=0\\ t&=\frac{18+\sqrt{18^{2}+4(4.905)(25)}}{2(4.905)}=4.744\ \text{s} \end{aligned}

Range

R=vxt=24(4.744)=113.86 mR=v_x t=24(4.744)=113.86\ \text{m}

Answer: time of flight t≈4.74t\approx 4.74 s; horizontal range R≈113.9R\approx 113.9 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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