Question

A baseball player is 98m away from home plate when he throws the ball at a 30 degree angle above horizontal from a height of 1.45m (the height of his hand).

a) What is the initial speed of the throw of the ball to land exactly on home plate? In order to solve this problem, set up equations for x and y position vs time. Put in all known values. You will have two remaining unknowns: Vo (the magnitude of the initial velocity) and t. Solve the system of equations.

b) How long does it take the ball to get to home plate?

c) What is the maximum height that the ball is above the ground?

Answer 1

(a) Given

x = 98 m

= 30

y_o = 1.45 m

(a) Using kinematics

y_f = y_o + v_yt + 1/2a_yt²

0 = 1.45 + vsin * t - 4.9t² --------- (1)

In x direction we have

x = vcos * t

t = x / v cos

put this in first equation

0 = 1.45 + vsin * x / vcos - 4.9 * (x / vcos)²

0 = 1.45 + xtan - 4.9 * x² / v²cos²

0 = 1.45 + 98 * tan 30 - 4.9 * 98² / v² cos²30

0 = 58.0303 - 62746.13 / v²

v = 32.88 m/s

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(b) t = 98 / 32.88 * cos 30

t = 3.44 sec

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(c) at maximum height, v_y = 0

so,

v_y² - v_oy² = 2ah

h = v_oy² / 2g

h = (32.88 * sin 30 )² / 19.6

h = 13.78 m

height above the ground

H = 15.24 m