A tennis player hits a ball at ground level, giving it an initial velocity of 22
ID: 1686633 • Letter: A
Question
A tennis player hits a ball at ground level, giving it an initial velocity of 22.0 at 51.0 above the horizontal.What are the horizontal and vertical components of the ball's initial velocity? in m/s
How high above the ground does the ball go? in m
How long does it take the ball to reach its maximum height? in s
What is the ball's velocity at its highest point? in m/s
What is the ball's acceleration at its highest point? in m/s^2
For how long a time is the ball in the air? in s
When this ball lands on the court, how far is it from the place where it was hit? in m
Explanation / Answer
Hi, Given the initial velocity u = 22m/s, and angle of projection = 51deg with horizontal. The horizontal component of the initial velocity = ux = u * Cos(51) = 22 * 0.62932 = 13.85 m/s The vertical component of the initial velocity = uy = u * Sin(51) = 22 * 0.77715 = 17.097 m/s Maximum vertical height it will reach = hmax = (uy)^2 / 2*g = (17.097)^2 / 19.6= 14.91m. Time of flight = T = 2 * uy / g = 2 * 17.097 / 9.8 = 3.49sec. Hence time taken to reach maximum height = T / 2 = 1.245sec. At highest point the ball will not have any vertical velocity. i.e., its velocity at this point is totally its horizontal velocity = ux = 13.85 m/s At highest point the ball acc will be the acceleration due to gravity = 9.8 m/s Time of flight = time the ball in air = T = 2 * uy / g = 2 * 17.097 / 9.8 = 3.49sec. Time of flight = time the ball in air = T = 2 * uy / g = 2 * 17.097 / 9.8 = 3.49sec. Total horizontal distance travelled = Range = u^2 * Sin2? / g = 48.31m Hope this helps you. Hope this helps you.
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