Asked by Faria
Consider the motion of a projectile. It is fired at t = 0. Its initial speed is 17 m/s and its initial projection angle is 35◦ from the horizontal.
What is the maximum height of its tra- jectory? The acceleration due to gravity is 9.8 m/s2 .
Answer in units of m.
What is the maximum height of its tra- jectory? The acceleration due to gravity is 9.8 m/s2 .
Answer in units of m.
Answers
Answered by
Damon
vertical problem only
Vi = 17 sin 35 = 9.75
initial kinetic energy = (1/2) mVi^2
potential energy at top = m g h
so
g h = 1/2 Vi^2
h = VI^2/(2 g) = 9.75^2/(2*9.8) = 4.85
pathetic projectile :(
Vi = 17 sin 35 = 9.75
initial kinetic energy = (1/2) mVi^2
potential energy at top = m g h
so
g h = 1/2 Vi^2
h = VI^2/(2 g) = 9.75^2/(2*9.8) = 4.85
pathetic projectile :(
Answered by
R_scott
the initial vertical velocity is ... 17 m/s * sin(35º)
h = -4.9 t^2 + [17 * sin(35º)] t
max h is on the axis of symmetry ... t = [-17 sin(35º)] / (2 * -4.9)
h = -4.9 t^2 + [17 * sin(35º)] t
max h is on the axis of symmetry ... t = [-17 sin(35º)] / (2 * -4.9)
Answered by
Faria
thank you guys!
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