Question

Prove that there are two values of time for which a projectile

Answer 1

Answer:

We know that for projectile motion, the equation for the y-co-ordinate for the projectile is given by

We know that for projectile motion, the equation for the y-co-ordinate for the projectile is given byy=usin(θ)t−12gt2

We know that for projectile motion, the equation for the y-co-ordinate for the projectile is given byy=usin(θ)t−12gt2If we solve this for t, using the quadratic formula, we get

We know that for projectile motion, the equation for the y-co-ordinate for the projectile is given byy=usin(θ)t−12gt2If we solve this for t, using the quadratic formula, we gett=usin(θ)±u2sin2(θ)−2gy√g

Without loss of generality, let t1=usin(θ)g

+u2sin2(θ)−2gy√g

+u2sin2(θ)−2gy√gand t2=usin(θ)g−u2sin2(θ)−2gy√g

+u2sin2(θ)−2gy√gand t2=usin(θ)g−u2sin2(θ)−2gy√gNow, all that’s left to do is t1+t2.

+u2sin2(θ)−2gy√gand t2=usin(θ)g−u2sin2(θ)−2gy√gNow, all that’s left to do is t1+t2.The second terms simply cancel out.

+u2sin2(θ)−2gy√gand t2=usin(θ)g−u2sin2(θ)−2gy√gNow, all that’s left to do is t1+t2.The second terms simply cancel out.Of course, there is a special case where the second term is zero. This happens if the projectile is at the top of its trajectory. Then t1=t2and there aren’t two distinct times, but they do add up to the time of flight.

+u2sin2(θ)−2gy√gand t2=usin(θ)g−u2sin2(θ)−2gy√gNow, all that’s left to do is t1+t2.The second terms simply cancel out.Of course, there is a special case where the second term is zero. This happens if the projectile is at the top of its trajectory. Then t1=t2and there aren’t two distinct times, but they do add up to the time of flight.Also, if you equate the stuff under the radical sign to zero and solve for y, you simply get the maximum height the projectile goes to!

+u2sin2(θ)−2gy√gand t2=usin(θ)g−u2sin2(θ)−2gy√gNow, all that’s left to do is t1+t2.The second terms simply cancel out.Of course, there is a special case where the second term is zero. This happens if the projectile is at the top of its trajectory. Then t1=t2and there aren’t two distinct times, but they do add up to the time of flight.Also, if you equate the stuff under the radical sign to zero and solve for y, you simply get the maximum height the projectile goes to!Simpler method:

+u2sin2(θ)−2gy√gand t2=usin(θ)g−u2sin2(θ)−2gy√gNow, all that’s left to do is t1+t2.The second terms simply cancel out.Of course, there is a special case where the second term is zero. This happens if the projectile is at the top of its trajectory. Then t1=t2and there aren’t two distinct times, but they do add up to the time of flight.Also, if you equate the stuff under the radical sign to zero and solve for y, you simply get the maximum height the projectile goes to!Simpler method:Rearranging, we get

+u2sin2(θ)−2gy√gand t2=usin(θ)g−u2sin2(θ)−2gy√gNow, all that’s left to do is t1+t2.The second terms simply cancel out.Of course, there is a special case where the second term is zero. This happens if the projectile is at the top of its trajectory. Then t1=t2and there aren’t two distinct times, but they do add up to the time of flight.Also, if you equate the stuff under the radical sign to zero and solve for y, you simply get the maximum height the projectile goes to!Simpler method:Rearranging, we gett2−2ugsinθ+2yg=0

+u2sin2(θ)−2gy√gand t2=usin(θ)g−u2sin2(θ)−2gy√gNow, all that’s left to do is t1+t2.The second terms simply cancel out.Of course, there is a special case where the second term is zero. This happens if the projectile is at the top of its trajectory. Then t1=t2and there aren’t two distinct times, but they do add up to the time of flight.Also, if you equate the stuff under the radical sign to zero and solve for y, you simply get the maximum height the projectile goes to!Simpler method:Rearranging, we gett2−2ugsinθ+2yg=0Thus, comparing to (x−a)(x−b)=x2−(a+b)x+ab, we ca say that the sum of the roots is 2ugsin