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A ball thrown up is caught by the thrower after 4 s. How high did it go and with what velocity was it thrown?

a. 40 m, 10 m/s
b. 20 m, 10 m/s
c. 40 m, 20 m/s
d. 20 m, 20 m/s

User Gerri
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1 Answer

4 votes

Final answer:

The ball reached a height of 78.4 meters and was thrown with an initial velocity of -39.2 m/s (upward direction). Therefore, option (B) is correct.

Step-by-step explanation:

To find the height the ball reached and its initial velocity, we can apply the principles of projectile motion. When a ball is thrown up and caught by the thrower, it follows a vertical trajectory. The height it reached can be determined using the formula:

h = ut + (1/2)gt^2

where h is the height, u is the initial velocity, t is the time, and g is the acceleration due to gravity. In this case, the ball is caught after 4 seconds, so we can plug in the values:

h = ut + (1/2)gt^2

h = (0)t + (1/2)(9.8)(4)^2

Simplifying, we get:

h = (1/2)(9.8)(16)

h = 78.4 m

Therefore, the ball reached a height of 78.4 meters.

To find the initial velocity, we can use the equation:

v = u + gt

where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time. In this case, the final velocity is 0 m/s (since the ball is caught), and the time is 4 seconds. Plugging in the values, we get:

0 = u + (9.8)(4)

Solving for u, we get:

u = -9.8(4)

u = -39.2 m/s

Therefore, the ball was thrown with an initial velocity of -39.2 m/s (negative indicates upward direction).

User Shaju
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