physics problems?

Question

how do you find an object's initial velocity if only an initial high, the range and the above horizonal angle are given?
For example, if an object is fired at 10m above the ground with the angle of 40 degrees obove the horizonal and lands 15m away? what is its initial velocity?

Answer 1

The answer involves a bit of algebra and is easy to lose track of where you are.

Conservation of energy says that

0.5*m*v^2*sin(q)^2 + m*g*h = m*g*H

where H is the height at the peak of the object's travel, q is the initial angle, h is the starting height, v is the magnitude of the velocity, and g is the gravitation acceleration. Also, we know that

(1): H - h = 0.5*g*T1^2

where T1 is the time is takes to reach the top of its travel. This expression was obtained based on the fall from the peak to the starting height h. Using this expression and the conservation of energy expression

(2): T1 = v*sin(q)/g

The remaining distance h is covered in a time T2 given by

(3): h = v*sin(q)*T2 + 0.5*g*T2^2

Given the range of travel (L) we know that

(4): v*cos(q)*T = L

and the relation among the individual times is

(5): T = 2*T1 + T2

T1 is counted twice for the upward and downward motion of the arc. That is the time to reach the top of its travel from its initial height and to return to that height..

Using expression (5) and replacing T2 in expression (3) we have

(6): 0.5*g*(T^2 + 4*T1^2 - 4*T1*T) + v*sin(q)*(T - 2*T1) - h = 0

From equation (2) we can substitute for T1

(7): 0.5*g*(T^2 + 4*v^2*sin(q)^2/g^2 - 4*v*sin(q)/g * T) + v*sin(q)*(T -2*v*sin(q)/g) - h = 0

Simplifying this becomes

(8): 0.5*g*(T^2 - 4*v*sin(q)/g*T) + v*sin(q)*T - h = 0

Using Equation (4) to replace T

(9): 0.5*g*(L^2/(v^2*cos(q)^2) - 4*sin(q)*L/(cos(q)*g)) + L*sin(q)/cos(q) - h = 0

which, after rearranging, becomes

(10): g*L^2/2 - (h*cos(q)^2 + L*sin(q)*cos(q))*v^2 = 0

or

v^2 = g*L^2/(2*(h*cos(q)^2 + L*sin(q)*cos(q)))

Given the values you mentioned above, v = 9.125 m/s.

Answer 2

Backtrack through the problem

Y-Velocity = Initial-Velocity * sin(angle)

0 = Initial-Height + Y-Velocity * Time + .5 * Acceleration-of-Gravity * Time

Solve the quadratic for the Time.

And then solve for the Range

X-Velocity = Initial-Velocity * cos(angle)

Range = X-Velocity * Time

Answer 3

The initial velocity is = zero cuz it was at rest when it was fired.
and if the object stops then too the final velocity is zero cuz finally it was at rest .
that is the correct answer trust me!!!! im saying that cuz i cannot find a better way to say it. hope you get my point.
read the chapter once again and you will understand better.