Question

A shot-putter throws the shot with an initial speed of 13 m/s at an angle of 42 degrees above the horizontal. Assuming that the shot is released from a height of 2.1 meters above the ground, calculate the horizontal distance the shot travels before it lands.

Answer 1

Answer:

hi

Explanation:

The formulas for the components Vx and Vy of the velocity and components x and y of the displacement are given by

Vx = V0 cos(θ) Vy = V0 sin(θ) - g t

x = V0 cos(θ) t y = V0 sin(θ) t - (1/2) g t2

In the problem V0 = 20 m/s, θ = 25° and g = 9.8 m/s2.

The height of the projectile is given by the component y, and it reaches its maximum value when the component Vy is equal to zero. That is when the projectile changes from moving upward to moving downward.(see figure above) and also the animation of the projectile.

Vy = V0 sin(θ) - g t = 0

solve for t

t = V0 sin(θ) / g = 20 sin(25°) / 9.8 = 0.86 seconds

Find the maximum height by substituting t by 0.86 seconds in the formula for y

maximum height y (0.86) = 20 sin(25°)(0.86) - (1/2) (9.8) (0.86) 2 = 3.64 meters

b) The time of flight is the interval of time between when projectile is launched: t1 and when the projectile touches the ground: t2. At t = t1 and t = t2, y = 0 (ground). Hence

V0 sin(θ) t - (1/2) g t2 = 0

Solve for t

t(V0 sin(θ) - (1/2) g t) = 0

two solutions

t = t1 = 0 and t = t2 = 2 V0 sin(θ) / g

Time of flight = t2 - t1 = 2 (20) sin(θ) / g = 1.72 seconds.