Question

a ball is thrown upward with initial velocity of 20 m/s. (a) how long is the ball in the air? (b) what is the greatest height reached by the ball? (c) when is the ball 15 m above the ground?

Answer 1

a) t = 4.08 s

b) Response = 20.39m

c) t = 0.991 and 3.087 seconds.

Initially, the vertical speed was 20 m/s.

(a) how long is the ball in the air?

The ball is tossed upward, rises to its highest point, and then hits the ground again.

Consequently, the end height will be equal to the starting height.

h = h o = Om

Use the equation s= ut + -at.

2764 – An = 14

Put a negative sign for g because the ball is being thrown upwards and g is acting downwards.

Om = t - 0.5 * 9.81 m/s * 20 m/s

0 = 20 *t - 0.5 * 9.81 *t

20 *t=0.5*9.81 *t

20 = 0.5 * 9.81 *t

20 0.5*9.81

Answer (a): t = 4.08 s

(b) what is the greatest height reached by the ball?

Vertical velocity for the ball decreases to zero when it reaches its highest point.

Use the equation 2-u=295

-2ghmas for - u

(20m/s - (Om/s)2) = -2 * 9.81 m/s2 *

02 – 20= -2*9.81 *

-20% = -2 *9.81 *

202 = 2 * 9.81 *  

b) RESPONSE: H = 20.39m

(c) when is the ball 15 m above the ground?

The ball will be 15 meters above the earth on two separate occasions.

1. When climbing

2. When descending

Use the equation s= ut + -at.

2,264 – fn = 4

0.5 * 9.81 m/s * 15m = 20 m/s*t

15 = 20 *t - 0.5 * 9.81 *t

15 = 20t – 4.9057

4.905 – 20t + 15 = 0

Calculator-based quadratic equation solution

(c) t = 0.991 and 3.087 seconds.

Hence the answers are,

a) t = 4.08 s

b) RESPONSE: Hmar = 20.39m

c) t = 0.991 and 3.087 seconds.

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