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If an object is thrown in an upward direction from the top of a building 160 ft high at an initial speed of 21point82 mph what is its final speed when it hits the ground?
Mizyangel16
g = 9.81 m/s^2 , 160 feet = 48.768 metres, 21.82 mph = 9.754 metres/sec.
it passes top of building on way down at same speed so :
u = 9.754 m/s
v = ?
g = 9.81 m/s^2
s = 48.768 metres
if : v = square root of (u^2 +( 2 * g * s))
then : v = 32.434 m/s ( 72.55 mph )
Wiki User
∙ 12y ago
The final speed of the object when it hits the ground can be calculated using the kinematic equation for motion: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity, and s is the distance fallen. Given that the object is thrown upwards, its initial velocity will be negative. By plugging in the values, we can solve for the final velocity.
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