If a ball is kicked with an initial speed of 35 m/s at an angle of 45° to the horizon off a 70 m cliff.?

To answer how long it will take to hit the ground, you only need to consider the y-component of displacement of the ball (i.e. its speed in the x-direction doesn't affect how long it takes to fall vertically).

The y-direction displacement is given by:

y = y0 + v*t + (1/2)*a*t^2

Where y is the final y position, y0 is the initial y position, v is the initial velocity in the y direction (which is 35*sin(45), the y-component of the initial speed given), a is acceleration in the y direction (gravity in this case) and t is time.

If you assume that upwards is the position direction, downwards is negative, and that the ball starts at +70m and finishes at 0m, then you can re-write the equation as:

0 = 70 + (35*sin(45))*t - (1/2)*(9.81)*t^2

And now this is just a quadratic equation which you can solve (for t) using the quadratic formula (or an online calculator if you're lazy like me), the two solutions are:

t = -2 or 7 seconds

(negative time doesn't make sense, so the answer is 7 seconds)

To find the final kinetic energy you can just say (due to conservation of energy) that it will be equal to the initial kinetic energy plus the potential energy from being on the cliff. Otherwise, you can just calculate the final y-velocity, and use the final y-velocity and x-velocity (which is the same as the initial x-velocity) to calculate the final kinetic energy. In either case, you need to know the mass o the ball to get an actual number though.

The angle as it hits the ground is a little trickier. You need to find the final y-velocity using:

vf = vi + a*t

vf is the final y-velocity, vi is the initial y velocity (35sin(45)), a is acceleration (gravity) and t is time again. You know the initial velocity and acceleration, and you know (from the previous part) the time it takes to fall, so you can calculate the final velocity:

vf = 35*sin(45) - (9.81)*(7) = -44 m/s

And now, you can calculate the angle by using trigonometry with both the final x and y velocities, you need to draw a picture of it for it to make sense though. I can't draw one here, but the angle would be:

tan(θ) = vy / vx

θ = arctan( 44 / 35*cos(45) ) = 61 degrees to the horizontal (actually if you do this number, you get a negative answer, but you can ignore the negative sign).

Horizontal component,Vxi of the initial velocity of the ball = 35*sq rt(2), which will remain constant throughout till it hits ground. So we have Vxi = Vxf = 35*sq rt(2) = 49.50m/s----------- 1

Vertical component,Vyi of the initial velocity of the ball = 35*sq rt(2), which will change at the rate of -9.8 m/s^2.

Time to reach maximum height = (35/9.8)*sq rt(2) s = 3.57*sq rt(2) s = 5.05 s----------------------- 2

Maximum height, h reached above that of cliff is given by

h = [[35*sq rt(2)]^2]/(2*9.8) = 125 m

Time, T required for the ball to come down from h above the cliff to ground is given by

T = sq rt[{2(125+70)}/9.8] = 6.31 s -------------------------------------------------------------------------- 3

Answer to

Q.1 How long will it take for the ball to hit the ground?

From 1 and 3 , Total time = t+T = 5.05 +6.31 = 11.36 s

Q.2 What will the kinetic energy of the ball be as it hits the ground?

Vertical component, Vyf of the ball when it hits ground = 9.8*6.31 = 61.84 m/s

kinetic energy of the ball when it hits ground = 0.5*m*[Vxi^2 + Vyf^2] = 3137*m J

Q. 3 What will be the angle that the ball hits the ground?

Angle with horizontal will be = tan^-1[61.84/49.50] = 51.3 degree