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Stolb23 [73]
3 years ago
14

Suppose you "dig a hole" 52 feet deep and place a baseball machine in the hole, so that the balls fire out at 96 ft /s and land

on the ground outside the hole. What is the velocity of one of these baseballs when it hits the ground?
Physics
1 answer:
jok3333 [9.3K]3 years ago
3 0

Answer:

76.73 ft/s

Explanation:

Let the final velocity is v.

initial velocity, u = 96 ft/s

g = 32 ft/s²

height, h = 52 feet

use third equation of motion

v² = u² - 2 gh

v² = 96 x 96 - 2 x 32 x 52

v = 76.73 ft/s

Thus, the speed of the ball as it reaches the ground is 76.73 ft/s.

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The x-axis is "time".  OK.  You know that as the graph moves from left to right, it shows what's happening as time goes on.

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The total electric field consists of the vector sum of two parts. One part has a magnitude of E1 = 1300 N/C and points at an ang
V125BC [204]

Answer:

2954.6 N/C, 46.36 degree from positive  axis

Explanation:

E1 = 1300 N/C, θ1 = 35 degree

E2 = 1700 N/C, θ2 = 55 degree

Now write the electric fields in vector form

E1 = 1300 ( Cos 35 i + Sin 35 j) = 1064.9 i + 745.6 j

E2 = 1700 ( Cos 55 i + Sin 55 j) = 975.08 i + 1392.6 j

Resultant electric field

E = E1 + E2

E =  1064.9 i + 745.6 j + 975.08 i + 1392.6 j

E = 2039.08 i + 2138.2 j

Magnitude of E

E = sqrt (2039.08^2 + 2138.2^2)

E = 2954.6 N/C

Let it makes an angle Φ from X axis

tan Φ = 2138.2 / 2039.08 = 1.049

Φ = 46.36 degree from positive X axis.

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^ What he said

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