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Lady_Fox [76]
3 years ago
15

A boy throws a ball upward with a velocity of 4.50 m/s at 60.0o. What is the maximum height reached by the ball?

Physics
2 answers:
sammy [17]3 years ago
6 0

Answer:

0.774m

Explanation:

The formula for maximum height is given by:

hmax = ∨₀² ₓ Sin (α)² / 2 × g

where;

∨₀ = initial velocity

Sin (α) = angle of launch

g = gravitational acceleration which is equal to 9.8m/s²

Plugging in our values, we will have:

hmax = (4.50m/s)² × (Sin 60.0)² / 2 × 9.8m/s²

hmax= 20.25m/s × 0.75 / 19.8m/s²

hmax = 15.1875 / 19.8

hmax = 0.774m

Cloud [144]3 years ago
3 0

Answer:

3.1m

Explanation:

Since we only care about the y direction we only need to find vy. Once u draw your vector you will realize that vy= 4.5sin60=3.897m/s.

use vf²=v²+2a(y)

At the maximum height the velocity is 0 and since the object is in freefall, a=-g

Plug in all values

0=15.1875-2*9.8(y)

solve for y

-15.1875*2/-9.8=y

y=3.1m

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