How Does Projectile Motion Affect a Ball Thrown from a Cliff?

[Naeem]

Q. A man on the edge of a cliff H = 40 m high throws a ball directly upward. It returns past him 1.8 s later. (H actually is the height of the point of release of the ball above the base of the cliff. Neglect air resistance.) (Hint - Gravity causes a downward acceleration at the rate g = 9.81 m/s2.)

a) With what initial speed did the man throw the ball?

Ans. H = 40m

t = 1.8 seconds

Using,

x-x0 = v0t + 1/2 at^2

To find vo:


x -xo = 0 ( I wonder why x -xo , is zero,) when the height of the cliff is given to be 40m.

0 = 1.8u + 4.9 * (1.8)^2

Solving for u, we get:

v = 8.82 m/s, which is correct I think!

b) How high above the point of release did the ball go?

We can use,

v^2 = v0^2 + 2g (x-xo)

v = final velocity = 0 m/s
v0 from part a, is 8.82 m/s

and g = 9.81 m/s2

Plugging in all that, we get:

x-x0 = 3.964 m

c) What is the speed of the ball when it passes him on the way back down?

Same as a) 8.82 m/s

d) How long after the man throws it does the ball hit the ground at the bottom of the cliff?

This is where I am stuck...

Which equation to use?

v^2 = u^2 + 2g ( x - xo)

or

x - xo = v0t + 1/2 gt^2

and what would my variables be

v0 = 8.82 m/s

g = 9.81 m/s2

We need to find 't' and add it to the 1.7seconds in the problem

That would be the total time, to hit the ground,


Plz. help!

 

[whozum]

A man on the edge of a cliff H = 40 m high throws a ball directly upward. It returns past him 1.8 s later. (H actually is the height of the point of release of the ball above the base of the cliff. Neglect air resistance.) (Hint - Gravity causes a downward acceleration at the rate g = 9.81 m/s2.)

a)With what initial speed did the man throw the ball? Correct

The ball underwent constant acceleration for time = 1.8s.

x-x_0 = v_0t + 1/2 gt^2
x-x_0 = v_0(1.8) + 9.8*1.8^2/2

x-x_0 = 0 since there is no total displacement, this is because the signs of v_0*t and 1/2 gt^2 are opposite.

0 = v_0(1.8) - 15.876
v_0 = 15.876/1.8 = 8.82m/s

b) How high above the point of release did the ball go? Correct

v_0 = 8.82
h = h
g = 9.8

h = \frac{v_0^2}{2g} = \frac{77.79}{19.6} = 3.969m

c) Correct
d) How long after the man throws it does the ball hit the ground at the bottom of the cliff?
there's two ways you can do this, you can take it as freefall from the point where its right infront of him, or freefall from the highest point. The second one is much easier and we'll do that, the ball goes 3.969m above where he threw it, for a total height of 43.969m from the ground.

h = v_0t + \frac{gt^2}{2}

v_0t = 0 because at the top of the throw the ball is not moving.
h = \frac{gt^2}{2}

From here we solve for time:

t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{87.938}{9.8}} = 2.995s plus the time for it to get from his hand to the peak = 0.9s.

Total time = 3.89s

Method 2:

h = v_0t + \frac{gt^2}{2}

Solved for time via quadratic formula:

t = \frac{2v+2\sqrt{v^2+2gh}}{2g}

t = \frac{2(8.82) + 2\sqrt{8.82^2 + 2(9.8)(40)}}{2(9.8)} = 3.895s

 

[thepatientmental]

To find the time it takes for the ball to hit the ground, we can use the equation:

x-x0 = v0t + 1/2 gt^2

where x-x0 is the distance the ball has fallen, v0 is the initial velocity (which we found to be 8.82 m/s in part a), g is the acceleration due to gravity (9.81 m/s2), and t is the time it takes for the ball to hit the ground.

Since we know that the ball was thrown from a height of 40 m, we can set x-x0 equal to -40 m (since the ball is now below the point of release).

-40 = (8.82)(t) + 1/2 (9.81)(t)^2

Solving for t using the quadratic formula, we get two answers: t = 3.225 seconds or t = -2.441 seconds. Since we are looking for a positive time, we can disregard the negative answer. Therefore, it takes approximately 3.225 seconds for the ball to hit the ground. Adding this to the 1.8 seconds it took for the ball to reach its maximum height and return to the man, we get a total time of 5.025 seconds.

So, the man threw the ball with an initial speed of 8.82 m/s, it reached a maximum height of 3.964 m above the point of release, and it took a total of 5.025 seconds for the ball to hit the ground.