A ball was thrown up into the air. The ball traveled a total distance of 125m. The time taken was 11s. What is the average speed of the ball? What is the actual speed of the ball, assuming that g (accelaration due to gravity is present)? What is the value of this acceleration?
2007-05-06 04:37:19 · 9 answers · asked by netwiyan 1 in Science & Mathematics ➔ Physics
If the total distance the ball traveled was 125m, the maximum height was 125/2 (up and down). At the apex , the velocity was zero. Compute acceleration due to gravity from that point.
The previous answers are correct for average speed, but they fail to answer the rest of the question.
2007-05-06 05:00:51 · answer #1 · answered by Anonymous · 0⤊ 0⤋
This ball was NOT thrown upwards from the Earth! So call the gravitational acceleration gp for whatever planet it was done on.
The average speed was 125/11 m/s = 11.3636... m/s.
Since h = 1/2 gp t^2, gp = 2h/t^2
where h is the maximum height and t is the time to that height, 11/2 s = 5.5 s. Thus:
gp = 125/(5.5)^2 = 4.13223... m/s^2.
The initial speed is given by v_o^2 = 2 gp h = 125*125/(5.5)^2, so
v_o = 125/5.5 m/s = 22.7272... m/s.
This is just twice the average speed, of course, a result which depends on the fact that v(t) is a LINEAR function of t. (It wouldn't be true for other more general dependences!) The formula for the general velocity (sign convention with positive value upwards) is:
v = v_o - gp t = 125/5.5 - (125/5.5^2) t
= [125/5.5^2)] (5.5 - t) m/s, or numerically 4.13223...(5.5 - t) m/s.
Live long and prosper.
2007-05-06 07:20:08 · answer #2 · answered by Dr Spock 6 · 0⤊ 1⤋
ignoring the peak of the ball at launch (probable approximately 2 meters) Vf^2 = Vi^2 + 2 a x (15 m/s)^2 - 2 x (-9.8 m/s^2) x 10 m = Vi^2 225 m^2/s^2 + 196 m^2/s^2 = Vi^2 = 421 m^2/s^2 so Vi = 20.5 m/s of direction in case you had to comprise the two meters then 10 m could fairly be 8 m and the respond could be 19.5 m/s so say the respond is approximately 20 meters / 2nd replace: dm_cork you forgot the two whilst multiplying v^2 = u^2 + (2) x as Edward. S = Vot + a million/2 a t ^2. you forgot the Vot term.
2016-10-14 22:18:22 · answer #3 · answered by virgin 4 · 0⤊ 0⤋
The average speed is the distance divided by the time. The actual speed depends on where in the flight path the ball is at a given time. Due to gravity, the upward speed will slow as the energy is transferred to the pull of gravity and it will increase as the ball falls back to earth.
2007-05-06 04:53:42 · answer #4 · answered by fangtaiyang 7 · 1⤊ 1⤋
the original speed of ball=50m/s
accln=9.8 or 10m/s2
2007-05-06 05:01:36 · answer #5 · answered by Jan 2 · 1⤊ 0⤋
Speed=about 11.36 meters per second.
2007-05-06 04:53:40 · answer #6 · answered by Mr. Awesome 2 · 1⤊ 1⤋
9.8 meters per second^2
2007-05-06 05:04:24 · answer #7 · answered by Anonymous · 0⤊ 0⤋
average = 125/11 = 11.363636...m/s
2007-05-06 04:45:30 · answer #8 · answered by The Ponderer 3 · 1⤊ 1⤋
who cares
2007-05-06 05:09:12 · answer #9 · answered by gwen a 1 · 0⤊ 1⤋