Asked by Rodrigo
A ball thrown into the air lands on the same horizontal level, 31 m away, and 2.65 s later. Find the magnitude of the initial velocity.
Answers
Answered by
Steve
if it rose and fell for 1.325s, then it rose and fell 4.9*1.325^2 = 8.6m.
Now, for a given angle θ and initial velocity v,
range R = v^2 sin(2θ)/g
max height H = v^2 sin^2(θ)/g
R/H = 4cotθ
So, here, 31/8.6 = 4cotθ
cotθ = 0.90
θ = 48°
using the range R,
31 = v^2 sin96°/g
v^2 = 31g/sin96° = 305.5
v = 17.48
Now, for a given angle θ and initial velocity v,
range R = v^2 sin(2θ)/g
max height H = v^2 sin^2(θ)/g
R/H = 4cotθ
So, here, 31/8.6 = 4cotθ
cotθ = 0.90
θ = 48°
using the range R,
31 = v^2 sin96°/g
v^2 = 31g/sin96° = 305.5
v = 17.48
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.