MA99 Posted October 10, 2016 Report Share Posted October 10, 2016 Hi! im wondering how to solve these projectile motion problems: A particle is launched horizontally with a speed of 8.0 m s−1 from a point 20 m above the ground. a Calculate the time when the particle lands on the ground. b Determine the speed of the particle 1.0 s after launch. c Find the angle between the velocity and the horizontal 1.0 s after launch. d Determine the velocity with which the particle hits the ground. A soccer ball is kicked so that it has a range of 30 m and reaches a maximum height of 12 m. Determine the initial velocity (magnitude and direction) of the ball Reply Link to post Share on other sites More sharing options...
mushroom Posted October 10, 2016 Report Share Posted October 10, 2016 Ahhh I love projectiles... Okay, so in order to solve a, what you do is firstly divide your knowns into vertical and horizontal components. What you see when you do that is that the time needed for the projectile to finish it's horizontal path is the same as the time needed for it to finish it's vertical path. With that idea, you solve for time using your knowns in the vertical component: vi = 8.0 m/s , d = 20 m , a = -9.81 m/s2 , t = ? With these, you can calculate time using the formula d = vi x t + 0.5at2 and you should get 2.99 s For b, you need the speed after 1.0 s launch. In order to find this, you take inventory of your knowns (in y component): t = 1.0 s , vi = 0.0 m/s , a = -9.81 m/s2 , vf = ? You then use the formula a = (vf - vi) / t and end up with -9.81 m/s The direction of the velocity would be found simply using trig. You know that your angular speed would be below the horizontal, and is -9.81 m/s^2. You know your horizontal speed is 8.0 m/s to begin with. You therefore make a vector diagram showing this and calculate the angle using sine, in this case. You should get 35.4 degrees below the horizontal. The impact velocity would be found using conservation of energy. Assuming the energy is conserved and knowing you begin with both Ek and Ep but end with only Ek, Initial Ek + Initial Ep = Final Ek , 0.5mvi2 + mgh = 0.5mvf2 You know vi and can cancel m, therefore you can calculate vf, which should be 21.4 m/s Hope this helps! Reply Link to post Share on other sites More sharing options...
kw0573 Posted October 10, 2016 Report Share Posted October 10, 2016 (edited) 4 hours ago, mushroom said: vi = 8.0 m/s , d = 20 m , a = -9.81 m/s2 , t = ? With these, you can calculate time using the formula d = vi x t + 0.5at2 and you should get 2.99 s Minor corrections: Since you have defined up to be positive direction, d is actually -20m because it would be 20m below your starting point. That is, d has same sign as a in this case. 4 hours ago, mushroom said: The impact velocity would be found using conservation of energy. Assuming the energy is conserved and knowing you begin with both Ek and Ep but end with only Ek, Initial Ek + Initial Ep = Final Ek , 0.5mvi2 + mgh = 0.5mvf2 You know vi and can cancel m, therefore you can calculate vf, which should be 21.4 m/s Because part d) asks for velocity (hence direction), it would be easier to use vfy2-viy2 = 2ad, take vfy < 0, and do the trig as shown in b) and c). Note viy = 0 5 hours ago, MA99 said: A soccer ball is kicked so that it has a range of 30 m and reaches a maximum height of 12 m. Determine the initial velocity (magnitude and direction) of the ball So by saying t = time to reach max height and equation time in x direction to time in y direction. You want to eventually derive this: https://en.wikipedia.org/wiki/Projectile_motion#Relation_between_horizontal_range_and_maximum_height. After getting angle, use vfy2-viy2 = 2ad to solve for viy,which is vsin (theta), and then solve for initial speed v. Edited October 11, 2016 by kw0573 1 Reply Link to post Share on other sites More sharing options...
mushroom Posted October 11, 2016 Report Share Posted October 11, 2016 Oh hehe, I didn't notice those, nor did I even notice the last question. Thanks! Reply Link to post Share on other sites More sharing options...
MA99 Posted October 11, 2016 Author Report Share Posted October 11, 2016 Thank you so much! Reply Link to post Share on other sites More sharing options...
adityamenon10 Posted April 3, 2020 Report Share Posted April 3, 2020 On 10/11/2016 at 2:22 AM, mushroom said: Ahhh I love projectiles... Okay, so in order to solve a, what you do is firstly divide your knowns into vertical and horizontal components. What you see when you do that is that the time needed for the projectile to finish it's horizontal path is the same as the time needed for it to finish it's vertical path. With that idea, you solve for time using your knowns in the vertical component: vi = 8.0 m/s , d = 20 m , a = -9.81 m/s2 , t = ? With these, you can calculate time using the formula d = vi x t + 0.5at2 and you should get 2.99 s For b, you need the speed after 1.0 s launch. In order to find this, you take inventory of your knowns (in y component): t = 1.0 s , vi = 0.0 m/s , a = -9.81 m/s2 , vf = ? You then use the formula a = (vf - vi) / t and end up with -9.81 m/s The direction of the velocity would be found simply using trig. You know that your angular speed would be below the horizontal, and is -9.81 m/s^2. You know your horizontal speed is 8.0 m/s to begin with. You therefore make a vector diagram showing this and calculate the angle using sine, in this case. You should get 35.4 degrees below the horizontal. The impact velocity would be found using conservation of energy. Assuming the energy is conserved and knowing you begin with both Ek and Ep but end with only Ek, Initial Ek + Initial Ep = Final Ek , 0.5mvi2 + mgh = 0.5mvf2 You know vi and can cancel m, therefore you can calculate vf, which should be 21.4 m/s Hope this helps! Could you explain why angular velocity must be -9.81 and not vertical velocity? i seem to be stumped... Reply Link to post Share on other sites More sharing options...
kw0573 Posted April 5, 2020 Report Share Posted April 5, 2020 On 4/3/2020 at 10:20 AM, adityamenon10 said: Could you explain why angular velocity must be -9.81 and not vertical velocity? i seem to be stumped... a is acceleration Reply Link to post Share on other sites More sharing options...
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