Asked by Anonymous
A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 11.0 m/s when the hand is 1.50 m above the ground.
How long is the ball in the air before it hits the ground? (The student moves her hand out of the way.)
v(final)=v(initial)+at
initial velocity is 15 and final velocity is zero. acceleration is 9.8 right.
how does height fit into this equation. isn't that necessary too.
can u please help me. i keep getting .738s as my answer
How long is the ball in the air before it hits the ground? (The student moves her hand out of the way.)
v(final)=v(initial)+at
initial velocity is 15 and final velocity is zero. acceleration is 9.8 right.
how does height fit into this equation. isn't that necessary too.
can u please help me. i keep getting .738s as my answer
Answers
Answered by
drwls
height (y) vs time is given by
y = 1.50 + 11 t - 4.9 t^2.
Solve for the time (t) when y = 0. You will have to solve a quadratic equation. Take the positive one of the two answers.
You don't use the velocity vs. time equation in this problem.
y = 1.50 + 11 t - 4.9 t^2.
Solve for the time (t) when y = 0. You will have to solve a quadratic equation. Take the positive one of the two answers.
You don't use the velocity vs. time equation in this problem.
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