Answer:
The take-off speed is 41.48 [tex]\frac{m}{s}[/tex]
Explanation:
Given :
Range [tex]R = 77[/tex] m
Projectile angle [tex]\alpha =[/tex] 13°
From the formula of range,
[tex]R = \frac{v_{o} ^{2} }{g} \sin 2\alpha[/tex]
Find the velocity from above equation,
[tex]v_{o} = \sqrt{\frac{gR}{\sin 2 \alpha } }[/tex]
[tex]v_{o} = \sqrt{\frac{9.8 \times 77}{\sin 26} }[/tex] ( ∵ [tex]g = 9.8[/tex] [tex]\frac{m}{s^{2} }[/tex] )
[tex]v_{o} = 41.48[/tex] [tex]\frac{m}{s}[/tex]
Therefore, the take-off speed is 41.48 [tex]\frac{m}{s}[/tex]
