Dynamics in Two Dimensions Question

[Enduro]

Homework Statement

During baseball practice, you go up into the
bleachers to retrieve a ball. You throw the
ball back into the playing field at an angle of
42° above the horizontal, giving it an initial
velocity of 15 m/s. If the ball is 5.3 m above
the level of the playing field when you throw
it, what will be the velocity of the ball when
it hits the ground of the playing field?

<=42°
vi=15m/s
h=5.3m
vf=?

Homework Equations

h=-0.5gt^2+VosinθΔt

vf=vi+aΔt

The Attempt at a Solution

5.3m=-0.5(9.8)t^2+15(sin 42°)Δt
t=1.02s

vf=(15)(sin 42°)+(-9.8)(1.02)
vf=0.041m/s

vx^2 + vy^2 (square root)
(15 x cos 42)^2 + 0.041^2
=11.15m/s

what am i doing wrong? Answer is 18m/s.

 

[lewando]

Your problem is localized here:

5.3m=-0.5(9.8)t^2+15(sin 42°)Δt
t=1.02s

The result you got for t is not correct.

 

[PerryKid]

This looks more like projectile motion. In this case, your equation: 5.3m=-0.5(9.8)t^2+15(sin 42°)Δt is not necessary.

It seems to be more useful as $Y=Y_{o}+V_{oy}t-\frac{gt^{2}}{2}$.

 

[haruspex]

h=5.3m

It starts at 5.3m above the final height, so what is the change in height?

 

[Nickie]

Your approach is correct, but there is a small error in your calculation. When calculating the final velocity, you need to use the value of the acceleration due to gravity as -9.8 m/s^2, not just -9.8. This is because the acceleration is acting downwards, so it should be negative.

Corrected calculation:

vf=(15)(sin 42°)+(-9.8)(1.02)
vf=11.15 m/s

Therefore, the final velocity of the ball when it hits the ground will be 11.15 m/s.