A baseball is thrown at an angle of 25degrees relative to the ground at a speed of 23.0 m/s. If the ball was?

Question

caught 42.0m from the thrower, how long was it in the air? How high was the tallest spot in the ball's path?

Answer 1

We will resolve the speed in to
horizontal and vertical components
V horiz=23*cos25Deg
V vert.=23*sin25Deg
Time in air=42/23*cos25
=42/23*.9063
=42/20.84
=2.01sec
v^2-u^2=2gs
v=0
u=23*sin25deg
g=9.81m/sec^2
s=Height to Highest Point
=-u^2/2g
=-[23*sin25]^2/2*9.81
=-[23*.4226]^2/2*9.81
=-9.72*9.72/2*9.81
=-4.81m
{negative sign indicates
against gravity}

Answer 2

The horizontal component of the initial velocity does not change. It is v0x = v0*cos(T) (T=25º, v0=23m/sec). The ball will travel a distance v0x*t. Equate that to 42m and calculate t.

The equation for the ball's height is v0y*t - .5*g*t^2,, where g = accell of gravity (9.8m/sec^2). v0y is the vertical component of initial velocity, v0y = v0*sin(T). In this case, however, you use half the t computed above, because the ball reaches max height halfway through its trajectory (based on level ground and symmetry).

Answer 3

For you lazy mfs -4.81 is the answer. (Negative means against gravity)