Solving Projectile Motion Equations for Initial Velocity and Angle

[jumbo123]

Homework Statement

A ball is thrown horizontally with speed u from a height of 20m. It hits the ground with a speed 3u. Find the value of u, and the angle at which the ball strikes the ground.

Homework Equations

The equations of motion.

The Attempt at a Solution

I know that I must deal separately with the horizontal and vertical components of the motion.
I have worked out that the final vertical velocity will be 19.8m/s but I don't where to go next to find u.

 

[gneill]

Write an expression for the speed at which it hits the ground. Set it equal to the required condition.

 

[jumbo123]

I have realized that I now have a triangle where the x component is u, y component is 19.8 and the hypotenuse is 3u. So I could easily work out u and then the angle by using simple trig.

Is this correct?

 

[gneill]

Looks good! Perhaps consider using Pythagoras.

 

[rwisz]

I would approach this problem by using the equations of motion for projectile motion. The horizontal and vertical components of the motion can be treated separately, as they are independent of each other.

To find the initial velocity, u, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration (in this case, due to gravity), and t is the time the object is in motion. Since we know the final velocity (3u) and the acceleration (9.8 m/s^2), we can plug those values into the equation and solve for u.

3u = u + (9.8)(t)
2u = (9.8)(t)
u = (9.8)(t)/2

Next, we can use the equation y = ut + (1/2)at^2 to find the time, t, that the object is in motion. Since we know the initial height (20m) and the final height (0m), we can set y = -20m (taking downward as the positive direction), and solve for t.

-20 = (9.8)(t)^2/2
t = √(40/9.8) = 2.02 seconds

Now that we have the time, we can plug that back into the equation for u to find its value.

u = (9.8)(2.02)/2 = 9.9 m/s

To find the angle at which the ball strikes the ground, we can use the equation tanθ = vy/vx, where θ is the angle, vy is the final vertical velocity, and vx is the final horizontal velocity. We already know vy (19.8 m/s) and vx (3u = 3(9.9) = 29.7 m/s), so we can plug those values in and solve for θ.

tanθ = (19.8)/(29.7)
θ = tan^-1(19.8/29.7) = 33.7°

Therefore, the initial velocity of the ball is 9.9 m/s and it strikes the ground at an angle of 33.7°.