A projecttile has inital velocity 50m/s at angle 60 degreeabove the horizontal.

Question

A projecttile has inital velocity 50m/s at angle 60 degreeabove the horizontal. exactly 1.5s later, find (a) the rate atwhich the height is increasing, (b) the speed, that is, the rate atwhich (diagonal) distance is being traversed along the curved path,(c) the angle between the direction of motion and the horizontal.Find also (d) how long after the inital instant the highest pointis reached, (e) the velocity at the highest point, (f) the maximumheight, (g) the accleration at the highest point. I got this question wrong on the quiz, the professor did notshow us how to get the answer. Please help, i want to learn from mymistake and not make the same mistake on the midterm. A projecttile has inital velocity 50m/s at angle 60 degreeabove the horizontal. exactly 1.5s later, find (a) the rate atwhich the height is increasing, (b) the speed, that is, the rate atwhich (diagonal) distance is being traversed along the curved path,(c) the angle between the direction of motion and the horizontal.Find also (d) how long after the inital instant the highest pointis reached, (e) the velocity at the highest point, (f) the maximumheight, (g) the accleration at the highest point. I got this question wrong on the quiz, the professor did notshow us how to get the answer. Please help, i want to learn from mymistake and not make the same mistake on the midterm.

Explanation / Answer

A 50 m/s at 60 degrees, we find the sin() for the height change:50*sin(60) = 43.3 m/s y = 43.3t - (1/2)(9.8)t^2 (equation ofmotion, google if you want the derivation) y' = 43.3 - 9.8t (equation for velocity) Plug in t = 1.5, and you'll get: y' = 28.6 m/s B 50m/s * cos(60) = 25 m/s C I might be misunderstanding this question, but if you mean atthe beginning, isn't it already given at 60 degrees? D The highest point is reached when dy/dt = 0 y' = 43.3 - 9.8t = 0 t = 4.41s E At the highest point, there's no up-down velocity, so it's allsideways; because there's no acceleration/deceleration on thehorizontal component, it's just our 50cos(60) = 25. F We know the time it takes to get to the maximum height,so: y = 43.3t - (1/2)(9.8)t^2 where t = 4.41: 95.7 m G The acceleration at the highest point is the same as it is atany point: 9.8 m/s^2 (or 9.80665 depending on what school/countryyou're in/who your professor is). -- I'm pretty sure that's right. Hope that helps.

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