Science, asked by kitkaaaaatt, 10 hours ago

A broad-jump athlete takes off at an angle of 30° above the horizontal and reaches the 200 m away.

a. What was her initial velocity?
b. How high did she jump?
c. How long was she in the air?

Answers

Answered by avnimittal1428
1

Answer:

Projectile motion concept.

R= u²sin2θ/ g

Given,

R= 200m

θ= 30°

By placing the values,

200= u²×√3/2(sin 2θ = 60°=√3/2) /9.8

=u= √200×9.8×2/√3

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