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
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
Similar questions