Question #4581

Natalya Lisovskaya set a track and field record in 1987 when she put the shot so that its horizontal displacement was 22.63 m. Assume that the angle of the initial velocity of the shot was such that it maximized the range, and for simplicity, assume the height of the ball when it landed was the same as the height at which it left her hand. If she were to put the shot vertically with the same speed, what is the maximum height above her hand that the shot would reach?

Expert's answer

The angle of the initial velocity of the shot such that it maximize the range is pi/4. Let's find the initial speed of a shot.

Let V1 and V2 be the vertical and the horizontal speed components. As the angle is pi/4, we see that V1 = V2 = V/sqrt(2).

Horizontal: distance D = V2*t = V*t/sqrt(2). (*)

Vertical: V1*t = (gt^2)/2, so, V*t/sqrt(2) = (gt^2)/2, and we see that D = (gt^2)/2. So, t = sqrt(2D/g) = sqrt(2*22.63/9.8) = 2.14 seconds.

From (*): the initial speed V = D*sqrt(2)/t = 22.63*sqrt(2)/2.14 = 14.95 m/s.

Let's find the maximum height.

Vt = (gt^2)/2, so, V = gt/2 and t = 2V/g = 3.05 seconds.

V - gt = 0 (0 is the speed at the top point), so time to get to the top point is t = V/g = 14.95/9.8 = 1.53 seconds.

Height H = Vt - (gt^2)/2 = 14.95*1.53 - (9.8*1.53^2)/2 = 11.40 meters.

Let V1 and V2 be the vertical and the horizontal speed components. As the angle is pi/4, we see that V1 = V2 = V/sqrt(2).

Horizontal: distance D = V2*t = V*t/sqrt(2). (*)

Vertical: V1*t = (gt^2)/2, so, V*t/sqrt(2) = (gt^2)/2, and we see that D = (gt^2)/2. So, t = sqrt(2D/g) = sqrt(2*22.63/9.8) = 2.14 seconds.

From (*): the initial speed V = D*sqrt(2)/t = 22.63*sqrt(2)/2.14 = 14.95 m/s.

Let's find the maximum height.

Vt = (gt^2)/2, so, V = gt/2 and t = 2V/g = 3.05 seconds.

V - gt = 0 (0 is the speed at the top point), so time to get to the top point is t = V/g = 14.95/9.8 = 1.53 seconds.

Height H = Vt - (gt^2)/2 = 14.95*1.53 - (9.8*1.53^2)/2 = 11.40 meters.

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