Answer:
Explanation:
a) The formulas for the components Vx and Vy of the velocity and components x and y of the displacement are given by x = 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.
c) In part c) above we found the time of flight t2 = 2 V0 sin(θ) / g. The horizontal range is the horizontal distance given by x at t = t2.
range = x(t2) = V0 cos(θ) t2 = 2 V0 cos(θ) V0 sin(θ) / g = V02 sin(2θ) / g = 202 sin (2(25°)) / 9.8 = 31.26 meters
d) The object hits the ground at t = t2 = 2 V0 sin(θ) / g (found in part b above)
The components of the velocity at t are given by
Vx = V0 cos(θ) Vy = V0 sin(θ) - g t
The components of the velocity at t = 2 V0 sin(θ) / g are given by
Vx = V0 cos(θ) = 20 cos(25°) Vy = V0 sin(25°) - g (2 V0 sin(25°) / g) = - V0 sin(25°)
The magnitude V of the velocity is given by
V = √[ Vx2 + Vy2 ] = √[ (20 cos(25°))2 + (- V0 sin(25°))2 ] = V0 = 20 m/s