Question

A physics student throws a ball with initial speed V0 = 10 m/s at 45 degrees. The ball strikes the wall 3.0 m away

a. When will it strike the wall? Answer = .43s
b. Is the ball ascending/descending?
c. What is the height at which the ball strikes the wall? (With respect to the initial height) Answer: 2m
d. Find the impact velocity of the ball Answer: 7i+2.9j

Answer 1

I think his answer is wrong for a possibly.

\[Vy = 10\sin45, Vx = 10\cos45\]

So I used the equation:\[Y = Vyi(t)−1/2(g)(t^2)\]
\[3 = 10\sin45(t) -1/2(10)(t^2) = -5t^2+7.1t-3\]
but I got t = .34 s

Answer 2

Im not sure about d., does anyone know how he got 7i+2.9j?

Answer 3

They've given horizontal displacement as 3, not vertical. Use \(S=ut\) horizontally instead.

For d, horizontal component of the velocity will not change. Find the vertical velocity component at the impact by applying appropriate equations vertically.

Answer 4

a) you need to consider horizontal movement: \[v_x=v_0·\cos(\theta) \rightarrow s_x=v_0·\cos(\theta)·t \rightarrow t_{impact}=\frac{ 3 }{ 10·\cos(45)}=0.42 \sec\]b) Peak height should happen when:
\[v_y=v_0·\sin(\theta)-g t=0\rightarrow t_{PEAK}=\frac{ v_0·\sin(\theta) }{ g }=0.72\sec\]That means that before reaching peak height, it has hit the wall, therefore it was ascending
c)Height is given by:\[h(t)=v_0·\sin(\theta)·t-\frac{ 1 }{ 2 }g t^2\rightarrow h(t=0.42)=2.12m\]
d)Velocity is given by:
\[\bar v(t)=v_0·\cos(\theta) \hat i+[v_0\sin(\theta)-g·t] \hat j\] just replace t with the impact time (0.42 sec) and get the vector

Answer 5

Thanks! that makes sense