Task:
A football leaves the toe of a painter at 200 ft/sec. and strikes the ground 500 ft away.
a.) Determine its initial angle of projection
b.) Find the total time of flight
c.) Find the maximum height reached.
Solution:
v 0 = 200 f t s v _ {0} = 2 0 0 \frac {f t}{s} v 0 = 200 s f t R = 500 f t R = 5 0 0 f t R = 500 f t R = v 0 cos α ⋅ t R = v _ {0} \cos \alpha \cdot t R = v 0 cos α ⋅ t H = v 0 sin α ⋅ t 2 − g ⋅ ( t 2 ) 2 2 = v 0 2 sin 2 α 2 g H = v _ {0} \sin \alpha \cdot \frac {t}{2} - g \cdot \frac {\left(\frac {t}{2}\right) ^ {2}}{2} = \frac {v _ {0} ^ {2} \sin^ {2} \alpha}{2 g} H = v 0 sin α ⋅ 2 t − g ⋅ 2 ( 2 t ) 2 = 2 g v 0 2 sin 2 α sin 2 α = ( v 0 sin α ⋅ t 2 − g ⋅ ( t 2 ) 2 2 ) ⋅ 2 g v 0 2 = ( v 0 sin α ⋅ t − g ⋅ ( t 2 ) 2 ) ⋅ g v 0 2 \sin^ {2} \alpha = \left(v _ {0} \sin \alpha \cdot \frac {t}{2} - g \cdot \frac {\left(\frac {t}{2}\right) ^ {2}}{2}\right) \cdot \frac {2 g}{v _ {0} ^ {2}} = \left(v _ {0} \sin \alpha \cdot t - g \cdot \left(\frac {t}{2}\right) ^ {2}\right) \cdot \frac {g}{v _ {0} ^ {2}} sin 2 α = ( v 0 sin α ⋅ 2 t − g ⋅ 2 ( 2 t ) 2 ) ⋅ v 0 2 2 g = ( v 0 sin α ⋅ t − g ⋅ ( 2 t ) 2 ) ⋅ v 0 2 g t = R v 0 cos α t = \frac {R}{v _ {0} \cos \alpha} t = v 0 cos α R sin 2 α = ( v 0 sin α ⋅ R v 0 cos α − g ⋅ ( R v 0 cos α 2 ) 2 ) ⋅ g v 0 2 = \sin^ {2} \alpha = \left(v _ {0} \sin \alpha \cdot \frac {R}{v _ {0} \cos \alpha} - g \cdot \left(\frac {\frac {R}{v _ {0} \cos \alpha}}{2}\right) ^ {2}\right) \cdot \frac {g}{v _ {0} ^ {2}} = sin 2 α = ⎝ ⎛ v 0 sin α ⋅ v 0 cos α R − g ⋅ ( 2 v 0 c o s α R ) 2 ⎠ ⎞ ⋅ v 0 2 g = = ( sin α ⋅ R cos α − g ⋅ ( R v 0 cos α 2 ) 2 ) ⋅ g v 0 2 = ( sin α ⋅ R cos α − g ⋅ R 2 4 v 0 2 cos 2 α ) ⋅ g v 0 2 = = ( 4 v 0 2 ⋅ cos α ⋅ sin α ⋅ R − g ⋅ R 2 4 v 0 2 cos 2 α ) ⋅ g v 0 2 = g ⋅ R 4 v 0 4 cos 2 α ⋅ ( 4 v 0 2 ⋅ cos α ⋅ sin α − g ⋅ R ) \begin{array}{l}
= \left(\sin \alpha \cdot \frac {R}{\cos \alpha} - g \cdot \left(\frac {\frac {R}{v _ {0} \cos \alpha}}{2}\right) ^ {2}\right) \cdot \frac {g}{v _ {0} ^ {2}} = \left(\frac {\sin \alpha \cdot R}{\cos \alpha} - \frac {g \cdot R ^ {2}}{4 v _ {0} ^ {2} \cos^ {2} \alpha}\right) \cdot \frac {g}{v _ {0} ^ {2}} = \\
= \left(\frac {4 v _ {0} ^ {2} \cdot \cos \alpha \cdot \sin \alpha \cdot R - g \cdot R ^ {2}}{4 v _ {0} ^ {2} \cos^ {2} \alpha}\right) \cdot \frac {g}{v _ {0} ^ {2}} = \frac {g \cdot R}{4 v _ {0} ^ {4} \cos^ {2} \alpha} \cdot (4 v _ {0} ^ {2} \cdot \cos \alpha \cdot \sin \alpha - g \cdot R)
\end{array} = ( sin α ⋅ c o s α R − g ⋅ ( 2 v 0 c o s α R ) 2 ) ⋅ v 0 2 g = ( c o s α s i n α ⋅ R − 4 v 0 2 c o s 2 α g ⋅ R 2 ) ⋅ v 0 2 g = = ( 4 v 0 2 c o s 2 α 4 v 0 2 ⋅ c o s α ⋅ s i n α ⋅ R − g ⋅ R 2 ) ⋅ v 0 2 g = 4 v 0 4 c o s 2 α g ⋅ R ⋅ ( 4 v 0 2 ⋅ cos α ⋅ sin α − g ⋅ R ) sin 2 α = g ⋅ R 4 v 0 4 cos 2 α ⋅ ( 4 v 0 2 ⋅ cos α ⋅ sin α − g ⋅ R ) \sin^ {2} \alpha = \frac {g \cdot R}{4 v _ {0} ^ {4} \cos^ {2} \alpha} \cdot (4 v _ {0} ^ {2} \cdot \cos \alpha \cdot \sin \alpha - g \cdot R) sin 2 α = 4 v 0 4 cos 2 α g ⋅ R ⋅ ( 4 v 0 2 ⋅ cos α ⋅ sin α − g ⋅ R ) α 1 = 11.85 ∘ , α 2 = 78.15 ∘ \alpha_ {1} = 11.85{}^{\circ}, \quad \alpha_ {2} = 78.15{}^{\circ} α 1 = 11.85 ∘ , α 2 = 78.15 ∘ t 1 = R v 0 cos α 1 = 2.554 s t _ {1} = \frac {R}{v _ {0} \cos \alpha_ {1}} = 2.554 \, \text{s} t 1 = v 0 cos α 1 R = 2.554 s t 2 = R v 0 cos α 2 = 12.174 s t _ {2} = \frac {R}{v _ {0} \cos \alpha_ {2}} = 12.174 \, \text{s} t 2 = v 0 cos α 2 R = 12.174 s H 1 = v 0 2 sin 2 α 1 2 g = 26.232 ft H _ {1} = \frac {v _ {0} ^ {2} \sin^ {2} \alpha_ {1}}{2 g} = 26.232 \, \text{ft} H 1 = 2 g v 0 2 sin 2 α 1 = 26.232 ft H 2 = v 0 2 sin 2 α 2 2 g = 595.852 ft H _ {2} = \frac {v _ {0} ^ {2} \sin^ {2} \alpha_ {2}}{2 g} = 595.852 \, \text{ft} H 2 = 2 g v 0 2 sin 2 α 2 = 595.852 ft
**Answer:**
a) α 1 = 11.85 ∘ , α 2 = 78.15 ∘ \alpha_{1} = 11.85{}^{\circ}, \alpha_{2} = 78.15{}^{\circ} α 1 = 11.85 ∘ , α 2 = 78.15 ∘
b) t 1 = R v 0 cos α 1 = 2.554 s t_1 = \frac{R}{v_0\cos\alpha_1} = 2.554 \, \text{s} t 1 = v 0 c o s α 1 R = 2.554 s
t 2 = R v 0 cos α 2 = 12.174 s t _ {2} = \frac {R}{v _ {0} \cos \alpha_ {2}} = 12.174 \, \text{s} t 2 = v 0 cos α 2 R = 12.174 s
c) H 1 = v 0 2 sin 2 α 1 2 g = 26.232 ft H_{1} = \frac{v_{0}^{2}\sin^{2}\alpha_{1}}{2g} = 26.232 \, \text{ft} H 1 = 2 g v 0 2 s i n 2 α 1 = 26.232 ft
H 2 = v 0 2 sin 2 α 2 2 g = 595.852 ft H _ {2} = \frac {v _ {0} ^ {2} \sin^ {2} \alpha_ {2}}{2 g} = 595.852 \, \text{ft} H 2 = 2 g v 0 2 sin 2 α 2 = 595.852 ft 
