Question

In the Olympic shot-put event, an athlete throws the shot with an initial speed of 12.0m/s at a40.0? 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
?

Answer 1

Do it in three parts. In the first use the kinematic equation of motion to find the maximum height above the initial height as well as the time required to get there, then use the same equation to calculate the time to reach the bottom (max height + initial). In the third use the same equation, but now do it to calculate the length traveled over the sum of both times. See complete work below:

Equation for distance and height:
D = Vo*t + 1/2*a*t^2
V = Vo + a*t
a = -g (deceleration due to gravity)
Vo = 12*sin(40) (upward component of velocity)

To find time to maximum height solve V equation for t:
V = Vo + a*t = Vo - g*t = 0 (at top, it is instantaneously still)
t = (Vo - V)/g = (12*sin(40) - 0)/9.81
t = 0.786285sec

Now plug into distance equation to find total height above start:
D = Vo*t + 1/2*a*t^2
D = 12*0.786285 - 1/2*9.81*0.786285^2
D = 6.40293m

D_total = D + Do = 6.40293 + 1.80 = 8.20293

Now that you know the distance it has to fall, calculate its time with the same equation as before, but solve for t. Now the initial velocity is 0 as it was instantaneously still at the top and it is accelerating with gravity so a = g:
D = Vo*t + 1/2*a*t^2
t = -(Vo + (Vo^2 + 2*D*a)^(1/2))/a
and also
t = -(Vo - (Vo^2 + 2*D*a)^(1/2))/a
plug in for each:
t = -(0 + (0^2 + 2*8.20293*9.81)^(1/2))/9.81 = -1.2932 (not valid because negative)
t = -(0 - (0^2 + 2*8.20293*9.81)^(1/2))/9.81 = 1.2932

Now the total time in the air is the time reaching the top of the throw and then coming back, so just find the distance traveled horizontally in that time. The horizontal component of velocity is Vo = 12*cos(40). We assume there is no wind resistance so we assume it does not decelerate so a = 0:
t = t_up + t_donw = 0.786285 + 1.2932 = 2.0795
D = Vo*t + 1/2*a*t^2 = Vo*t + 1/2*0*t^2 = Vo*t
D = Vo*t = 12*cos(40)*2.0795 = 19.1159m

So it will travel 19.1 m in about 2.08 seconds