Projectile Motion Homework: Solving for Time, Position, and Velocity

[VaioZ]

Homework Statement

A ball is thrown vertically upward with an initial speed of 19.6 m/s from the top of a building 44.1 m tall. Assuming that there is no air resistance.
a.) At what two times is the ball 10.0 m above its starting point?
b.) What is the position of the ball at t = 5.00 s?
c.) How long will it take the ball to strike the ground?
d.) What is the velocity of the ball when it strikes the ground?

Homework Equations

Kinematics Equation/Projectile Equations

The Attempt at a Solution

a.) First I took 10 m as my delta Y then 19.6 m/s as may initial velocity to get the time
10 m = (19.6 m/s)(t) - (1/2)(9.8m/s^2)(t)^2
t = 3.3997 s

b.) First I took the time to get to the max height which is (19.6 m/s)/(9.8 m/s^2) = 2 s as my Tmax
then I compute for the Ymax which is Y=(19.6 m/s)(2 s) - (1/2)(9.8 m/s^2)(2 s)^2
Ymax = 19.6 m

then I added 19.6 m + 44.1 m = 63.7 m to get the height from base of the building to the max height of the projectile

then next is to get the position of the ball at 5 s
Y=(19.6 m/s)(5 s) - (1/2)(9.8 m/s^2)(2 s)^2
Y=-24.5 m

Then I substract 63.7 m - 24.5 m = 39.2 m

c.) 63.7 m = (19.6 m/s)(t) - (1/2)(9.8)(t)^2
but then it can't be solved sooooo I think I got wrong somewhere

pls help thank you.

 

[UMath1]

The displacement to the ground is relative to the starting position. In this case the starting position would be set as 44.1 m. So for part C, x=-44.1 m.

Alternatively, you could set 63.7 m as your starting position like you did. Then the position at the ground would be -63.7 m. But, the initial velocity would be zero since at the maximum height the ball has no velocity.

 

[VaioZ]

The displacement to the ground is relative to the starting position. In this case the starting position would be set as 44.1 m. So for part C, x=-44.1 m.

Alternatively, you could set 63.7 m as your starting position like you did. Then the position at the ground would be -63.7 m. But, the initial velocity would be zero since at the maximum height the ball has no velocity.

So you mean to say -63.7 m = (0)(t)-1/2(9.8)(t)^2 ?? I assume you're pertaining to c.)

and also is my a.) and b.) right? just to be clear.

 

[UMath1]

So for part A, you should have two times. The ball rises AND falls. I think the time you have is when the ball hits 10m as it is falling.
Heres a graph of displacement as a function of time for the situation:

For part B, I think all you need to is plug 5 in for t and solve. There is no need to calculate the maximum height.
Your new equation for part C is correct.

 

[VaioZ]

So for part A, you should have two times. The ball rises AND falls. I think the time you have is when the ball hits 10m as it is falling.
Heres a graph of displacement as a function of time for the situation: View attachment 87466

For part B, I think all you need to is plug 5 in for t and solve. There is no need to calculate the maximum height.
Your new equation for part C is correct.

So about part A, you mean I should multiply by 2 the t? 3.3997 multiply by 2? I thought the t for the ball rises and falls is the same by definition. Right?
Okayyy will solve later. I'll update you later for my final sol'n thank you!

 

[UMath1]

Not quite. If you look at the graph, the ball appears to be at y=10 at two points. One of them is the you discovered: 3.399. You have to find the other one which appears to be at around 0.6 seconds.

Think of it this way. You toss the ball up. At .6 seconds it reaches 10 m. It keeps going higher and starts to fall. On its descent it reaches the same point once again. But more time has passed. It is now 3.399 seconds since you threw it.

I suggest you factor the equation as a quadratic or use the quadratic formula.