A senior physics class conducting a research project on projectile motion constructs a device that can launch a cricket ball. The launching device is designed so that the ball can be launched at ground level with an initial velocity of 28 m/s at an angle of 30° to the horizontal.
a. Calculate the horizontal component of the velocity
b. Calculate the vertical component of the velocity of the ball:
C. At what time will the ball reach its maximum height?
D.What is the maximum height that is achieved by the ball?
Answers
Answer:
What error has been made in these calculations?
A senior physics class conducting a research project on projectile motion constructs a device that can launch a cricket ball. The launching device is designed so that the ball can be launched at ground level with an initial velocity of 28 m/s at an angle of 30 degrees to the horizontal.
At what time after being launched with the ball return to the ground?
2 * (0 - 14) / -9.8 = 2.9 s <== correct
What is the velocity of the ball as it strikes the ground?
v_vert = u + at = 0 + 9.8 * 2.9 = 28 m/s down
v = sqrt(v_vert^2 + [28cos(30 degrees)]^2) = 37 m/s <== incorrect
theta = atan[28cos(30 degrees) / v_vert] = 40 degrees <== incorrect
v = 37 m/s 140 degrees <== incorrect
Calculate the horizontal range of the ball.
28cos(30 degrees) * 2 * (0 - 14) / -9.8 = 69 m <== correct
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Can I do part b without having done part c, and if so, how?
A softball of mass 250 g is thrown with an initial velocity of 16 m/s at an angle theta to the horizontal. When the ball reaches its maximum height its kinetic energy is 16 J.
a. What is the maximum height achieved by the ball from its point of release?
x = 16 / 0.250 / 9.8 = 6.5 m <== correct
b. Calculate the initial vertical velocity of the ball.
u = 16sin(theta)
c. What is the value of theta?
0 = u^2 + 2ax
u = sqrt(-2ax)
theta = asin(sqrt(-2ax) / 16) = 45 degrees <== correct
(back to) b. u = 11 m/s <== correct.
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