A 0.23 kg rock is projected from the edge of the top of a building with an initi

Question

A 0.23 kg rock is projected from the edge of
the top of a building with an initial velocity of
11.5 m/s at an angle 46? above the horizontal.
The building is 7.67 m in height.At what horizontal distance, x, from the
base of the building will the rock strike the
ground? Assume the ground is level and
that the side of the building is vertical. The
acceleration of gravity is 9.8 m/s2.
Answer in units of m

3) A basketball player 2.01 m tall wants to
make a basket from a distance of 12.1 m. The
hoop is at a height of 3.05 m.If he shoots the ball (from a height of
2.01 m) at a 19.6?angle, at what initial speed
must he shoot the basketball so that it goes
through the hoop without striking the backboard? The acceleration due to gravity is
9.8 m/s2 Neglect air friction.
Answer in units of m/s

Explanation / Answer

for the first question, you do not need to take into consideration the mass of the rock because the gravity is a key bringing down the object.
initial velocity of a rock in x component: 11.5cos46m/s=8
of y component: 11.5 sin 46m/s= 8.27
x component velocity will not change.
we are looking for its distance traveled along the x axis, which is
vcos(t)= x

t= x/(vcos)we can say that y component traveled 7.67m downwards.

x= vt+ 1/2at^2

-7.67 = 8.27(x/8)+ 1/2(-9.8)(x/8)^2

x= -5.32 or 18.824, positive answer is 18.824

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