Kinematics: projectiles question

[lektor]

Given the context of this question being quite large and my main problem being the solving i won't bother with diagrams but anyway

the kinematics conditons are

Such that i have 10 = (Vcos15)t + 0.5gt^2
1.9 = (Vsin15)t + 0.5gt^2

Can anyone please show me the steps for solving this simultaneously for V,

Cheers.

 

[Fermat]

Write the 1st eqn as:Vcos15 = whatever
Write the 2nd eqn as :Vsin15 = whichever

Divide the 1st eqn by the 2nd eqn
Solve for t
substitute for t into the 1st or 2nd eqn
Solve for V

 

[kyle8921]

I understand your frustration with the large context of this question and the difficulty of solving it without diagrams. However, it is important to remember that kinematics is the study of motion without considering the forces that cause the motion, and therefore, diagrams may not always be necessary.

To solve this problem, we can use the equations of motion for a projectile, which are:
x = x0 + Vx0t + 0.5at^2
and
y = y0 + Vy0t + 0.5at^2

In this case, x0 = y0 = 0 (assuming the projectile is launched from ground level), Vx0 = Vcos15, and Vy0 = Vsin15.

We can substitute these values into the equations given in the problem:
10 = (Vcos15)t + 0.5gt^2
and
1.9 = (Vsin15)t + 0.5gt^2

Now, we have two equations with two unknowns (V and t). We can solve for t in one of the equations and substitute that value into the other equation to solve for V.

Solving for t in the first equation, we get:
10 = (Vcos15)t + 0.5gt^2
10 = (Vcos15)t + 0.5(9.8)t^2
0.5(9.8)t^2 + (Vcos15)t - 10 = 0

Using the quadratic formula, we can solve for t:
t = (-b ± √(b^2 - 4ac)) / 2a
where a = 0.5(9.8), b = Vcos15, and c = -10

Substituting these values, we get:
t = (-Vcos15 ± √(V^2cos^215 - 4(0.5(9.8))(-10))) / 2(0.5(9.8))

Simplifying, we get:
t = (-Vcos15 ± √(V^2cos^215 + 98)) / 9.8

Now, we can substitute this value of t into the second equation:
1.9 = (Vsin15)t + 0.5gt^2
1.9 = (