How do you solve these types of problems, where the information provided is very limited?:
- Starting from rest, a child throws a ball with an initial speed v at B degrees above the horizontal. The child then chases after the ball, accelerating at a constant acceleration a. If the child wants to catch the ball at the same height it was thrown, what must be the child’s acceleration a? Use g for the gravitational constant.
Since you have no numerical data, your final answer should look like a formula. You will need an expression for the expired time, t, when the ball has returned to that original height. So, you will need the vertical component of the initial velocity. I will call that Vvi.
Vvi = v*sin(B)
Declare the reference position for the vertical variable, y, to be the height from which it was thrown. Then an expression for the expired time, t, when the ball has returned to y=0, can be obtained using one of the familiar kinematic formulas.
y = Vvi*t + (1/2)*a*t^2
The acceleration will be -g, so the formula becomes
0 = v*sin(B)*t + (1/2)*(-g)*t^2
t = 2*v*sin(B)/g
During that time the ball will travel horizontally a distance, x, given by the product of that time and the horizontal component of v:
x = Vhi*t = v*cos(B)*2*v*sin(B)/g
Now the child needs to accelerate such that the ball is caught when it returns to y=0. Using
x = Vi*t + (1/2)*a*t^2
Now plug in what we know,
v*cos(B)*2*v*sin(B)/g = 0*t + (1/2)*a*(2*v*sin(B)/g)^2
Now simplify and solve for a
v*cos(B) = 0*t + (1/2)*a*(2*v*sin(B)/g)
a = 2*g*v*cos(B)/(v*sin(B))
a = 2*g*cot(B)
There were several steps there. I urge you to go thru my math to find any mistake I might have made. The dimensions are correct for an acceleration, so that's good but I have make an error before, could happen again.
I hope this helps,