The problem involves a ball thrown vertically upward with an initial velocity of 30 m/s, and participants are exploring how long it takes to reach the maximum height, as well as related calculations regarding height and velocity at different points in the trajectory.
Some participants have provided calculations and are questioning the correctness of their answers. There is a mix of confirmations and clarifications regarding the use of equations and the signs in calculations. Multiple interpretations of the problem are being explored, particularly regarding the height at which the speed equals half the initial speed.
Participants are working under the constraint of not using decimals in their final answers, and there is a focus on ensuring the correct application of kinematic equations.
Right!tommy2st said:0 is the final velocity. is the answer 3s? i think i may have got it
30=10t
t=3
anyone am i right?
Almost. Careful with signs.also the height it reaches using 3?
would i use d=30*3 + 1/2(10)(3^2)
Yep!tommy2st said:would it be d= 30*3 + 1/2(-10)(3^2)?
giving me 45?
Right. So you can find the distance using average speed X time.tommy2st said:average speed 15 m/s?
Solve it exactly as you solved the first part. First figure out the time.also i need help with this part... On the way up, at what height is the speed of the ball equal to half its speed of projection?
i get that it wants me to find at what distance v = 15 but I am having trouble munipulating the equations to solve this
Right. The initial velocity is 30 m/s, not 15. (That's the final velocity.)tommy2st said:okay thank you ill try and edit this
t=1.5
d=(15)(1.5)+(1/2)(-10)(1.5^2)= 11.25?
or (30)(1.5)+(1/2)(-10)(1.5^2)= 33.75?
i think its the second one