A football is kicked 62 m. If the ball is in the air for 6 s, (a) find its horizontal velocity Vx. Enter m/s as unit and use g = 10. m/s2.
(b) What is its initial vertical velocity Vy? Enter m/s as unit.
(c) With what initial speed was the football kicked? Enter m/s as unit.
The answers I got were... (a) either 2 m/s or 4 m/s ............ (b) 30 m/s ......... (c) I don't know really, it seems to me the same as what b asked for.
The computer keeps telling me I'm getting it wrong (it grades the three of them as a set, so even if two of them are right, if 1 of the answers is wrong, then it's ALL wrong; the thing is.... it doesn't tell me which answer is wrong... am I doing things right? Are my answers correct? If not, can someone PLEASE help me?
I PROMISE 10 POINTS FOR YOUR TIME AND EFFORT!
(b) What is its initial vertical velocity Vy? Enter m/s as unit.
(c) With what initial speed was the football kicked? Enter m/s as unit.
The answers I got were... (a) either 2 m/s or 4 m/s ............ (b) 30 m/s ......... (c) I don't know really, it seems to me the same as what b asked for.
The computer keeps telling me I'm getting it wrong (it grades the three of them as a set, so even if two of them are right, if 1 of the answers is wrong, then it's ALL wrong; the thing is.... it doesn't tell me which answer is wrong... am I doing things right? Are my answers correct? If not, can someone PLEASE help me?
I PROMISE 10 POINTS FOR YOUR TIME AND EFFORT!
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a) velocity = distance/time. You have the horizontal distance and the time it took to travel that distance. If you divide the given distance by the given time you aren't going to end up with either 2 m/sec or 4 m/sec.
Maybe you forgot that the horizontal motion is constant velocity. No acceleration. Don't use a = 10 m/sec^2 to describe this motion, it's just d = v * t.
b) That's correct. It takes 3 seconds for that vertical velocity to decrease to 0 m/sec, which means it takes 3 seconds to reach the top of the motion, so it takes another 3 seconds to come back down again. Total, 6 seconds.
c) You have the horizontal and vertical components of a vector. The total speed is the magnitude of the velocity vector, so it's sqrt(vx^2 + vy^2).
Maybe you forgot that the horizontal motion is constant velocity. No acceleration. Don't use a = 10 m/sec^2 to describe this motion, it's just d = v * t.
b) That's correct. It takes 3 seconds for that vertical velocity to decrease to 0 m/sec, which means it takes 3 seconds to reach the top of the motion, so it takes another 3 seconds to come back down again. Total, 6 seconds.
c) You have the horizontal and vertical components of a vector. The total speed is the magnitude of the velocity vector, so it's sqrt(vx^2 + vy^2).
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All projectile motion questions can be answered using these two parametric equations for horizontal and vertical.
Horizontal: X(t) = Vo t cos theta, Y(t) = Vo t sin theta - 1/2 g t^2
Where, Vo = initial speed, theta = angle of inclination with the ground, g = gravity and t = time
We are told that a ball is kicked 62 m. So, X(0) = 0 and X(t-final) = 62. We are also told that t-final = 6 s.
So, X(6) = Vo 6 cos theta = 62
Y(6) = Vo 6 sin theta - 1/2 10 * 6^2
You have 2 equations and 2 unknowns. Take it from here.
Horizontal: X(t) = Vo t cos theta, Y(t) = Vo t sin theta - 1/2 g t^2
Where, Vo = initial speed, theta = angle of inclination with the ground, g = gravity and t = time
We are told that a ball is kicked 62 m. So, X(0) = 0 and X(t-final) = 62. We are also told that t-final = 6 s.
So, X(6) = Vo 6 cos theta = 62
Y(6) = Vo 6 sin theta - 1/2 10 * 6^2
You have 2 equations and 2 unknowns. Take it from here.