A ball is projected with initial speed v0 at an angle θ from the ground level. After two seconds, the ball is displaced 40 m horizontally and 53 m vertically above its launch point... A ball is projected with initial speed v0 at an angle θ from the ground level. After two seconds, the ball is displaced 40 m horizontally and 53 m vertically above its launch point. Determine (a) the horizontal and vertical components of the initial velocity of the ball, (b) the values v0 and θ, (c) the range of the ball, and (d) the times after projection when the height of the ball is 30 m. Read more

Steps to Solve

1. Determine horizontal velocity component ($v_{0x}$)
   The horizontal displacement ($d_x$) after time $t$ can be calculated using the formula: $$ d_x = v_{0x} \cdot t $$ Given that $d_x = 40 , \text{m}$ and $t = 2 , \text{s}$, we can express $v_{0x}$ as: $$ v_{0x} = \frac{d_x}{t} = \frac{40, \text{m}}{2, \text{s}} = 20, \text{m/s} $$

2. Determine vertical velocity component ($v_{0y}$)
   The vertical displacement ($d_y$) can be expressed using the formula: $$ d_y = v_{0y} \cdot t - \frac{1}{2} g t^2 $$ Here, $d_y = 53 , \text{m}$ and $g \approx 9.81, \text{m/s}^2$. Plugging in the values: $$ 53 = v_{0y} \cdot 2 - \frac{1}{2} \cdot 9.81 \cdot (2)^2 $$ This simplifies to: $$ 53 = 2v_{0y} - 19.62 $$ Solving for $v_{0y}$: $$ 2v_{0y} = 53 + 19.62 = 72.62 $$ $$ v_{0y} = \frac{72.62}{2} = 36.31, \text{m/s} $$

3. Calculate initial velocity ($v_0$) and angle ($\theta$)
   The initial velocity $v_0$ can be calculated using: $$ v_0 = \sqrt{v_{0x}^2 + v_{0y}^2} $$ Substituting the values: $$ v_0 = \sqrt{(20)^2 + (36.31)^2} $$ Calculating the squares: $$ v_0 = \sqrt{400 + 1313.0561} $$ $$ v_0 \approx \sqrt{1713.0561} \approx 41.38 , \text{m/s} $$

4. Calculate the angle $\theta$
   The angle $\theta$ can be determined using: $$ \theta = \tan^{-1}\left(\frac{v_{0y}}{v_{0x}}\right) $$ Substituting the values: $$ \theta = \tan^{-1}\left(\frac{36.31}{20}\right) $$ Calculating: $$ \theta \approx \tan^{-1}(1.8155) \approx 61.57^\circ $$

5. Determine the range of the ball
   The range ($R$) can be calculated using: $$ R = v_{0x} \cdot t_f $$ Where $t_f$ is the total time of flight. The total time of flight can be calculated by finding the time to reach the peak and then the total duration using the vertical motion equations: $$ t_f = \frac{2v_{0y}}{g} $$ Substituting the values: $$ t_f \approx \frac{2 \cdot 36.31}{9.81} \approx 7.41 , \text{s} $$

Calculating the range: $$ R = 20 \cdot 7.41 \approx 148.2 , \text{m} $$

6. Determine the times when height is 30 m
   Using the vertical motion equation: $$ 30 = v_{0y} t - \frac{1}{2} g t^2 $$ Substituting: $$ 30 = 36.31t - \frac{1}{2}\cdot 9.81t^2 $$ This gives a quadratic equation: $$ 4.905t^2 - 36.31t + 30 = 0 $$

Solving using the quadratic formula: $$ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ Where:

• $a = 4.905$
• $b = -36.31$
• $c = 30$

Calculating the discriminant: $$ b^2 - 4ac = (-36.31)^2 - 4 \cdot 4.905 \cdot 30 $$ $$ = 1313.0561 - 588.6 \approx 724.4561 $$

Finding $t$ values: $$ t = \frac{36.31 \pm \sqrt{724.4561}}{2 \cdot 4.905} $$ This yields two times when height is 30 m.