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Pumpkin Toss In Denver, children bring their old jack-o-lanterns
to the top of a tower and compete for accuracy in hitting a target on the ground (FIGURE 4-21). Suppose that the tower is 9.0 m high and that the bull's-eye is a horizontal distance of 3.5 m
from the
launch point. If the pumpkin is thrown horizontally, what is the
launch speed needed to hit the bull's-eye?

Respuesta :

SaniShahbaz

Answer:

launch speed = v = 2.57 m/s

Explanation:

This problem is related to projectile motion. To find the launching speed of pumpkin, we first have to find the time of flight, then using R = vt, We can find launching speed of pumpkin.

Given data:

Height = h = 9.0 m

Horizontal distance = R = 3.5 m

Velocity = v = ?

As the pumpkin is thrown horizontally, So Initial vertical velocity of pumpkin is zero, So we can use 2nd equation of motion to find time of flight.

Using 2nd equationof motion

[tex]h = v_{i_{y}}t + \frac{1}{2}gt^{2}[/tex]

[tex]v_{i_{y}}[/tex] = 0

So,

[tex]h=\frac{1}{2}gt^{2}[/tex]

[tex]9=\frac{1}{2}(9.8)t^{2}[/tex]

t = 1.36 s

Using R = vt

v = R/t

v = 3.5/1.36

v = 2.57 m/s

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