In the olympic shotput event, an athlete throws the shot with an initial speed of 13.0 m/s at a 41.0 ∘ angle from the horizontal. the shot leaves her hand at a height of 1.80 m above the ground.(a) how far does the shot travel? (b) repeat the calculation of the first part for angle 42.5? .(c) repeat the calculation of the first part for angle 45 ? .(d) repeat the calculation of the first part for angle 47.5? .(e) at what angle of release does she throw the farthest?

(a) 18.93 m

First of all, we need to find the time it takes for the shot to reach the ground. This can be found by analyzing the vertical motion only, using the equation:

where

y(t) is the vertical position at time t

h = 1.80 m is the initial heigth

is the initial vertical velocity

t is the time

g = 9.8 m/s^2 is the acceleration of gravity

Substituting y(t)=0, we solve the equation for t, to find the time at which the shot reaches the ground:

which gives two solutions: t = -0.19 s and t = 1.93 s. We discard the former solution since it's negative, so the correct time is

t = 1.93 s

Now, the horizontal distance travelled by the shot is given by

where

is the horizontal velocity

t is the time

Substituting t = 1.93 s, we find the range of the shot:

(b) 18.96 m

We can solve the problem exactly in the same, replacing the previous angle with :

- The initial vertical velocity is

So the equation to find the time is

which gives t = 1.98 s as solution. The horizontal velocity is

And so the range of the shot is

(c) 18.84 m

Again, we solve using the new angle :

- The initial vertical velocity is

So the equation to find the time is

which gives t = 2.05 s as solution. The horizontal velocity is

And so the range of the shot is

(d) 18.70 m

Solving using the new angle ,

- The initial vertical velocity is

So the equation to find the time is

which gives t = 2.13 s as solution. The horizontal velocity is

And so the range of the shot is

(e) At

We see from the previous parts that the angle that gives the maximum range is . Therefore we can infer an important thing: while for a projectile launched from the ground (h=0) the maximum range is achieved when the angle of launch is , this is not true for a projectile launched from a height h.

Indeed, it can be demonstrated that for a projectile launched from height h, the angle at which the maximum range is achevied is given by

And indeed, by substituting u = 13.0 m/s and h = 1.80 m into the formula, we find

Which is in agreement with our findings.

b

step-by-step explanation: a third of 24 is 8, 8 times 3 is 24 so 8 - 2 would be 6

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