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A long jumper leaves the ground at an angleof 15.3â—¦ to the horizontal and at a speed of 11 m/s.
How far does he jump? The acceleration
due to gravity is 9.8 m/s
Answer in units of m.
2.What maximum height does he reach?
Answer in units of m.

Respuesta :

Ankit221005

Explanation:

Initial Velocity (u) = 11 m/sec

Angle of Projectile (theta) = 15.3°

(Ques - 1) How far does he jump?

(Ans - 1) Find Horizontal Range :

[tex]R = \frac{ {u}^{2} \sin(2 \alpha ) }{g} = \frac{(11)^{2} \sin(30.6°) }{9.8} [/tex]

[tex]Range = 6.17 \: m[/tex]

(Ques - 2) What maximum height does he reach?

(Ans - 2)

[tex]H = \frac{ {u}^{2} {sin }^{2} \alpha }{2g} = \frac{(11)^{2} {sin}^{2}15.3° }{19.6} [/tex]

[tex]Maximum \: Height = 0.41 \: m[/tex]

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