Can anyone help me with this physics problem?

Question

Snowballs are thrown with a speed of 13 m/s from a roof 7 m above the ground. Snowball A is thrown straight downward; snowball B is thrown in a direction 25 degrees above the horizontal. (A) Is the landing speed of snowball A greater than, less than, or the same as the landing speed of snowball B? Explain. (B) Verify your answer to part a by calculating the landing of both snowballs.

Answer 1

a) Conservation of energy dictates that they must be the same (ignoring air resistance).

b) initial PE + KE = mgh + 1/2 mvi^2
=
final KE (no PE) = 1/2 m vf^2

vf = sqrt (2gh + vi^2)

They give you h and vi. You know g. The angle is irrelevant for this calculation. Plugnchug.

Answer 2

*** Mistress Bekki is right, BUT, her answer wasn't very accurate with regards to certain things.

1)
The landing speed of snowball A will be greater than the landing speed of snowball B.


2)
Just calculate the final velocities with this equation:

vfy² = voy² + 2aΔy (note that this is exactly the same formula that Mistress Bekki made. Although she showed the formula, she obviously didn't use it)

Ball A:
vfy² = (13)² + 2(9.80)(7)
vfy = 17 m/s

For Ball B, to calculate the final velocity, we need the initial velocity. Since that ball was thrown with a trajectory, we must find the vertical component.

voy = vo*sin(ϑ) = 13*sin(25°) = 5.5 m/s

Ball B:
vfy² = (5.5)² + 2(9.80)(7)
vfy = 13 m/s

* EDIT: This is what I've figured out so far. The overall velocity will be the same (17.5 m/s), but the vertical velocities will be different.

Answer 3

Poo poo on the answer that poo poo'd Bekki.

The total energy for both balls when they are tossed is the same TE = PE + KE; where PE is the potential energy at start (they are both at h = 7 m above ground and have the same mass; so PE = mgh). And KE = 1/2 mU^2 for both; where U = 13 m/sec.

As there is no drag friction to do work on the balls as they travel and, therefore, change the make up, not the value, of TE, we can assert that TE = PE + KE = KE upon impact of both balls. PE = 0 because the balls are at ground level on impact. All of TE is converted to kinetic energy, but the total amount of energy remains at TE.

Therefore, TE = 1/2 mV^2 and V = sqrt(2TE/m) for both balls; where V is the impact velocity magnitude.

The only real difference between ball A and ball B is the makeup of V. In A V = vv, the vertical velocity as there is no horizontal velocity vh. For B, in vectors, we have V = vv + vh, but the magnitude of V is still the same for both balls. Only the direction of V is different.

Way to go, Bekki.

[NB: If there were drag friction, we'd have different results because drag is work done (W), but its a part of the energy budget; so that TE = W + PE + KE and, depending on a lot of variables not cited in this problem, the amount of work done on A could vary from the amount of work done on B. In which case the two KE's might not equal on impact.]

Answer 4

Ball A and Ball B hit the ground at the same speed.
(17.499 m/s.)

However, because ball A hits the ground vertically, 100% of it's kinetic energy is "normal", or perpendicular, to the ground.

Ball B, on the other hand, hits the ground at an angle of 47.677 degrees above the horizontal, and the vertical component(12.938m/s) is reduced by 26% as compared with ball A.

The area of impact is also larger for ball B, in effect spreading out the impact.

Answer 5

the landing speed of snowball A is greater.

split the projectile motion into vertical and horizontal components. then chuck away and ignore the horizontal.

the vertical componets of velocity will differ -
ball A 13m/s straight down
ball B = 13Sin(30) = 6.5m/s


therefore whn ball B falls back through 'start pos' on downwoard journey (ignoring air res) it will have velocity of 6.5m/s downwards. therefore half of ball A speed.

can't be bothered with part B, but just use your Eq of motion