Solve Kinematics Problem: Find θ for sx = symax

[FredericChopin]

Homework Statement

http://imgur.com/BPPbwpu

Homework Equations

s = ((v + u)*t)/2 - an equation of motion for constant acceleration, which is applicable in this situation.

The Attempt at a Solution

Let u be the velocity at which the projectile was launched.

It is given that s_x = s_ymax. So:

s_x = s_ymax = u*cos(θ)*t

Thus:

s_ymax/t = u*cos(θ)

Now, in the y-direction, the projectile was given initial velocity u*sin(θ), and at the top of its trajectory (where it has maximum height s_ymax), it has final velocity 0 m/s. Thus:

s_ymax = ((0 + u*sin(θ))*t)/2

, which means:

s_ymax/t = u*sin(θ)/2

However, since s_ymax/t = u*cos(θ) :

u*cos(θ) = u*sin(θ)/2

Thus:

cos(θ) = sin(θ)/2

2*cos(θ) = sin(θ)

2 = sin(θ)/cos(θ)

2 = tan(θ)

θ = tan^-1(2) = 63.4349°

However, this was marked wrong.

Can somebody see what could have gone wrong?

Thank you.

 

[ehild]

Homework Statement

http://imgur.com/BPPbwpu

Homework Equations

s = ((v + u)*t)/2 - an equation of motion for constant acceleration, which is applicable in this situation.

The Attempt at a Solution

Let u be the velocity at which the projectile was launched.

It is given that s_x = s_ymax. So:

s_x = s_ymax = u*cos(θ)*t

Thus:

s_ymax/t = u*cos(θ)

Now, in the y-direction, the projectile was given initial velocity u*sin(θ), and at the top of its trajectory (where it has maximum height s_ymax), it has final velocity 0 m/s. Thus:

s_ymax = ((0 + u*sin(θ))*t)/2

, which means:

s_ymax/t = u*sin(θ)/2

However, since s_ymax/t = u*cos(θ) :

u*cos(θ) = u*sin(θ)/2

Thus:

cos(θ) = sin(θ)/2

2*cos(θ) = sin(θ)

2 = sin(θ)/cos(θ)

2 = tan(θ)

θ = tan^-1(2) = 63.4349°

However, this was marked wrong.

Can somebody see what could have gone wrong?

Thank you.

The time to reach the highest point is not equal to the whole time of flight.

 

[FredericChopin]

The time to reach the highest point is not equal to the whole time of flight.

Ahh... I got it. It takes only half the total travel time to reach the top of the trajectory. Thank you.