A ball is thrown vertically upward with a speed of 1.53m/s...?

Question

A ball if thrown vertically upward with a speed of 1.53m/s from a point 4.21m above the ground. Calculate the time at which the ball will reach the ground. Include all the numbers and formulas you use to find the answer, even our teacher couldnt figure it out.

equations in their plain form would be nice, and the last answer is a little confusing, explain it a little.

Answer 1

Hmm... so
The ball goes up, then returnd to the same point and then it keeps on falling for 4.21m.

t(up) + t(down 1) + t(down 2)
t(up) = t(down 1)
V=gt
or V=gt(up)

t(up)=V/g=1.53/9.81=0.156 sec

t(down 2) is the time it takes to fall h=4.21m
let it be t
S=V0t + 0.5gt^2
gt^2 + 2V0t -2S=0

solving for t we have
t=0.784 s

So can we add ...

T(total)=0.156 +0.156 +0.784=1.096sec

Answer 2

For the ball to attain its optimal height, the formula is Vf - Vo = gT the place Vf = very final velocity = 0 (at optimal height) Vo = 490 cm/sec = 4.9 m/sec. g = acceleration as a results of gravity = 9.8 m/sec^2 (consistent) T = time Substituting values, 0 - 4.9 = (-9.8)T word the destructive sign related to the acceleration as a results of gravity. This basically denotes that the ball is slowing down because it is going up. fixing for T, T = 4.9/9.8 T = 0.5 sec. ( the ball will attain its optimal height at the instant) examine: Use the formula S = VoT - (a million/2)gT^2 to make certain the definitely optimal height of the ball the place S = optimal height attained by potential of ball and each and all the the remainder of the words have been previously defined. Substituting values, S = 4.9(0.5) - (a million/2)(9.8)(0.5^2) S = a million.225 m word -- The ball will under no circumstances attain the optimal height of 122.5 m (as given on your project). it may in basic terms bypass as severe as a million.225 m. there became probable a typo on your posting.

Answer 3

I agree with the idea of the answer, but I believe it should be

D = 4.21 Vo= -1.53 T ? D = VoT + 1/2 G(T) (T)

4.21 = -1.53 (T) + 1/2(9.8)(T) (T)

4.21+1.53 (T ) = 4.9 (T) (T)
4.21 + 1.53 = 4.9 (T) * cancel T on both side
work in this case

T = (4.21 + 1.53) / 4.9 = 1.17 sec.

because the distance travelled and acceleration are in the same direction (down) and initial velocity is up.

Answer 4

D = 4.21 Vo= 1.53 T ? D = VoT + 1/2 G(T) (T)

4.21 = 1.53 (T) + 1/2(9.8)(T) (T)

4.21-1.53 (T ) = 4.9 (T) (T)
4.21 - 1.53 = 4.9 (T) * cancel T on both side
work in this case

T = (4.21 - 1.53)/ 4.89

Answer 5

If your teacher is that illiterate, you can submit anything that resembles 'solution' and get away with it. There is no need to ask us.