已知初始速度与加速度，求解抛射小球的落地位置及偏角

Let's walk through solving this problem step by step:

Step 1: Determine when the ball lands #

The ball lands when its vertical (z-axis) position returns to 0. Using the kinematic equation for vertical motion:

• Initial z-velocity: $v_{0z} = 48$ ft/s
• z-acceleration: $a_z = -32$ ft/s²
• Position function: $z(t) = v_{0z}t + \frac{1}{2}a_z t^2$

Set $z(t) = 0$:

48t - 16t² = 0
t(48 - 16t) = 0

The non-trivial solution is $t = 3$ seconds (this is when the ball hits the ground).

Step 2: Calculate horizontal positions at landing #

East (x-axis) position #

No acceleration in the east direction, so:

• $x(t) = v_{0x}t = 40 * 3 = 120$ feet

South (y-axis) position #

Initial y-velocity is 0, with a constant southward acceleration:

• $y(t) = \frac{1}{2}a_y t^2 = 0.5*(-6)*(3)^2 = -27$ feet (the negative sign indicates south direction)

Step 3: Distance from origin #

The distance is the magnitude of the horizontal position vector:

d = sqrt(x² + |y|²) = sqrt(120² + 27²) = sqrt(15129) ≈ 123.0 feet

Step 4: Deflection angle from east #

Use trigonometry to find the angle θ between the east direction and the ball's landing position:

tanθ = |y| / x = 27 / 120 = 0.225
θ = arctan(0.225) ≈ 12.7 degrees

Final Results #

• Distance from origin: 123.0 feet
• Deflection angle south of east: 12.7 degrees

内容的提问来源于stack exchange，提问作者annac