An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal.
a) What is the maximum height reached by the object?
b) What is the total flight time (between launch and touching the ground) of the object?
c) What is the horizontal range (maximum x above ground) of the object?
d) What is the magnitude of the velocity of the object just before it hits the ground?
Answers
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
Given : An object is launched at a velocity of 36m/s in a direction making an angle of 29⁰ upward with the horizontal
To Find :
a) What is the maximum height reached by the object?
b) What is the total flight time (between launch and touching the ground) of the object?
c) What is the horizontal range
c) What is the magnitude of the velocity of the object just before it hits the ground?
Solution:
Horizontal Velocity = 20cos25° = 18.13 m/s
Vertical Velocity = 20Sin25° = 8.45 m/s
Horizontal Velocity remains constant
Vertical velocity becomes 0 at max height
V = u + at
a = - g = -9.8 m/s²
=> 0 =8.45 - 9.8t
=> t =0.86 sec
Total Time of flight = 2 * 0.86 = 1.72 sec
V² - U² = 2as
Max height = (v² - u²)/2a = -(8.45)²/(2*(-9.8))
= 3.64 m
Horizontal range = 1.72 x 18.13 = 31.18 m
magnitude of the velocity of the object just before it hits the ground
= 20 m/s ( based on conservation of energy )
Initial KE + initial PE = Final KE + Final PE
Initial and final PE = 0
Initial KE = Final KE
Hence Magnitude of velocity will be same
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