My problem is the following: King Arthur\'s knights fire a cannon from the top o

Question

My problem is the following:
King Arthur's knights fire a cannon from the top of the castlewall. The cannonball is fired at a speed of 56 m/s and an angle of35°. A cannonball that was accidentally dropped hits the moatbelow in 1 s.
How far from the castle wall does the cannonball hit theground?
What is the ball's maximum height above the ground?


My method is totally wrong or maybe I missed up along the line.Here it goes:
Part 1 question
I used the equation X(final) = X(initial) + V(initial)(change intime) + (1/2)(a)(change in time^2)
I plugged in the values : X(final) = .5(9.8 m/s^2)((1.00s)^2) = 4.9m

Then, I use the equation X(final) = x(initial) + ((56 m/s)(sin 35deg))/(2*-9.8 m/s^2). My answer is 57.54 m
I think this answers the first question but I am not sure!!!

Part 2 question
I use the equation (V(final)-V(initial))/a = change in time
I plugged in the values: {(0m/s) - ((56m/s)(sin 35 deg))}/{(-9.8m/s^2)}. My answer came out to be 3.278 s

Then I used the equation X(final) = x(initial) + V(initial)(changein time) + (1/2)(a)(change in time^2). I rearrange the equation tobe Change in time = sqrt{(2X(final)/a}
I plugged in the values: sqrt{(2*56m/s(cos35deg))/9.8 m/s^2}
My answer came out to be 3.427 s
I add up the time: 3.278s + 3.427s = 6.705 s
And that answers the second part.

So what do you think. Did I mess up somewhere?

Explanation / Answer

part A Xfinal = Xinitial + Vinitial*t + 1/2*Ax*t*t so, in vertical direction, when launched ball hits ground,    0 = 4.9 + 56*sin35*t - 1/2*9.8*t*t    so, 4.9*t*t - 56sin35*t - 4.9 = 0    so, t = ?    now in this time distance of ball from wall =Vhorizontal * t = 56cos35*t = ? m This answers the first part Part B For this part use conservation of mechanical energy.    1/2*m*(56sin35)2 + m*g*4.9 = m*g*h     since the horizontal speed remainsunchanged.    so, h = 4.9 + (56sin35)2 / (2*9.8) m =

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.