Physics question for module 5 (1 Viewer)

[coolpankake]

Who can help me with this one please, thanks:

Find the initial angle if you are given the range who equals 70m and the time equals 2.3 seconds

 

[chilli 412]

if we assume the projectile was launched at ground level and landed at ground level,
time of flight = 2.3
time to reach max height = 2.3/2 = 1.15
vertical velocity at max height = 0 (because the projectile has to turn)
v = u +at
we know that initial vertical velocity is given by u*sin(a), where 'a' represents the initial angle
at max height: 0 = usin(a) + (-9.8)(1.15)
rearranging, we get: usin(a) = 11.27 which implies that u = (11.27)/sin(a)
horizontal range formula: s = u*cos(a)*t , where 'u*cos(a)' represents initial horizontal velocity
we know total range is 70m, and this occurs at the end of flight when t = 2.3
therefore: 70 = 2.3ucos(a)
dividing both sides by 2.3, we get (70/2.3) = ucos(a) which implies that u = (70/2.3)/cos(a)
now that we have 2 expressions for u, set both equal to each other
(11.27)/sin(a) = (70/2.3)/cos(a)
rearranging by cross multiplying, we get
11.27cos(a) = (70/2.3)*sin(a)
dividing both sides by cos(a), we get
11.27 = (70/2.3)*tan(a)
dividing both sides by (70/2.3), we get
tan(a) = (25.921/70), now we can take inverse tangent of both sides and get
a = 20.3 degrees (1 d.p)
let me know if there were any mistakes because i had to type this out. process should be correct though

 

[coolpankake]

if we assume the projectile was launched at ground level and landed at ground level,
time of flight = 2.3
time to reach max height = 2.3/2 = 1.15
vertical velocity at max height = 0 (because the projectile has to turn)
v = u +at
we know that initial vertical velocity is given by u*sin(a), where 'a' represents the initial angle
at max height: 0 = usin(a) + (-9.8)(1.15)
rearranging, we get: usin(a) = 11.27 which implies that u = (11.27)/sin(a)
horizontal range formula: s = u*cos(a)*t , where 'u*cos(a)' represents initial horizontal velocity
we know total range is 70m, and this occurs at the end of flight when t = 2.3
therefore: 70 = 2.3ucos(a)
dividing both sides by 2.3, we get (70/2.3) = ucos(a) which implies that u = (70/2.3)/cos(a)
now that we have 2 expressions for u, set both equal to each other
(11.27)/sin(a) = (70/2.3)/cos(a)
rearranging by cross multiplying, we get
11.27cos(a) = (70/2.3)*sin(a)
dividing both sides by cos(a), we get
11.27 = (70/2.3)*tan(a)
dividing both sides by (70/2.3), we get
tan(a) = (25.921/70), now we can take inverse tangent of both sides and get
a = 20.3 degrees (1 d.p)
let me know if there were any mistakes because i had to type this out. process should be correct though

Thank you very much for the help. Let say they it was from a cliff. How will it change the answer then?

 

[chilli 412]

Thank you very much for the help. Let say they it was from a cliff. How will it change the answer then?

if it was from a cliff, we arent given enough information to solve. they would have to tell us what height the cliff is. from there you would follow a similar process EXCEPT if the projectile was launched from a cliff, the time for it to reach max height is not half of the total time of flight (because it continues to fall past its original vertical displacement as it falls down the cliff, adding more time to the total time of flight. you would have to mostly depend on s=ut+0.5at^2