PROJECTILE
A long-jumper leaves the ground at an angle of 20° above the horizontal at a speed of 11m/s.
Given:
- v¡ = 11 m/s
- θ = 20°
- g = 9.8 m/s²
1) The horizontal distance
[tex]R = \displaystyle\frac{v_{i}\:^{2} \sin2 \theta}{g}[/tex]
[tex]R = \displaystyle\frac{(11 \: m/s)^{2} \sin 40^{\circ}}{9.8 m/s^{2}}[/tex]
[tex]R = \displaystyle \tt 7.94 \: m[/tex]
Answer: [tex] \large \sf \bold{ \blue{7.94 \: m}}[/tex]
2) The maximum height reached
[tex]d_{y} = \displaystyle\frac{(v_{i} \sin \theta)^{2}}{2g}[/tex]
[tex]d_{y} = \displaystyle\frac{(11 \: m/s \times \sin 20^{\circ})^{2}}{2(9.8 m/s^{2})}[/tex]
[tex]d_{y} = \displaystyle \tt 0.72 \: m[/tex]
Answer: [tex] \large \sf \bold{ \blue{0.72 \: m}}[/tex]
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