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. Identify horizontal and vertical motions

   The horizontal motion has constant velocity, while the vertical motion is affected by gravity.

2. Calculate horizontal component of velocity

   Using the horizontal displacement: $$ x = v_{0x} \cdot t $$ where ( x = 40 , \text{m} ) and ( t = 2 , \text{s} ): $$ v_{0x} = \frac{x}{t} = \frac{40}{2} = 20 , \text{m/s} $$

3. Calculate vertical component of velocity

   Using the vertical displacement: $$ y = v_{0y} \cdot t - \frac{1}{2} g t^2 $$ where ( y = 53 , \text{m} ), ( g = 9.81 , \text{m/s}^2 ): $$ 53 = v_{0y} \cdot 2 - \frac{1}{2} \cdot 9.81 \cdot (2)^2 $$ Simplifying: $$ 53 = 2 v_{0y} - 19.62 $$ So, $$ 2 v_{0y} = 53 + 19.62 = 72.62 $$ Thus, $$ v_{0y} = \frac{72.62}{2} = 36.31 , \text{m/s} $$

4. Calculate initial speed and angle

   The initial speed ( v_0 ) is found using: $$ v_0 = \sqrt{v_{0x}^2 + v_{0y}^2} $$ Substitute the values: $$ v_0 = \sqrt{20^2 + 36.31^2} = \sqrt{400 + 1316.0561} $$ $$ v_0 \approx 42.64 , \text{m/s} $$

   To find angle ( \theta ): $$ \theta = \tan^{-1}\left(\frac{v_{0y}}{v_{0x}}\right) = \tan^{-1}\left(\frac{36.31}{20}\right) $$ Calculate ( \theta ): $$ \theta \approx 60.45^\circ $$

5. Calculate the range of the ball

   The range ( R ) is calculated using: $$ R = \frac{v_0^2 \sin(2\theta)}{g} $$ Substitute values: $$ R \approx \frac{(42.64)^2 \cdot \sin(120.9^\circ)}{9.81} $$ Simplifying gives: $$ R \approx 165.86 , \text{m} $$

6. Determine times when the height is 30 m

   Use the vertical motion equation: $$ 30 = v_{0y} t - \frac{1}{2} g t^2 $$ Rearranging gives a quadratic: $$ 0 = -4.905 t^2 + 36.31 t - 30 $$ Using the quadratic formula: $$ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ Here, ( a = -4.905, b = 36.31, c = -30 ): Discriminant: $$ b^2 - 4ac = 36.31^2 - 4 \cdot (-4.905) \cdot (-30) = 1316.0561 - 588.6 $$ Roots will be calculated using quadratic formula.