A ball is thrown upward with a velocity of 29.4 m/s from a tower that is 98 m high.?

Question

Calculate (a) the time it takes for the ball to reach its highest point; (b) the highest distance reached; (c) the distance and velocity after t = 2 s. and (d) the time it takes to reach the ground.

Answer 1

The equation to find the time to reach apogee is
v(t)=0=v0-g*t

using g=9.81
t=29.4/9.81
t=3.00 seconds

To find the height

y(t)=y0+v0*t-.5*g*t^2
y(3)=98+29.4*3-.5*9.81*9
y(3)=142 m

at t=2
y(2)=98+29.4*2-.5*9.81*4
137 m

v(2)=29.4-9.81*2
v(2)=9.78 m/s


to reach the ground
y(t)=0=98+29.4*t-.5*9.81*t^2

This quadratic has a positive and a negative root. The positive root is of interest
t=8.38 seconds
If you wonder why there is a negative root, that is the point in "time" where the ball would be thrown upward from the ground with an initial velocity that would result in the ball reaching y(0)=98m and v(0)=29.4 m/s

j

Answer 2

Use the formulas s=1/2at^2 and v=at for this problem.

time to apex = v/g = 29.4 / 9.8 = 3sec. (answer for a)

height of apex = 1/2at^2 + initial_height
0.5*9.8*3^2 + 98m = 44.1m + 98m
= 142.1m (answer to b)

Time to hit the ground from apex:
s=1/2at^2
142.1 = 0.5 * 9.8 * t^2
142.1 = 4.9 * t^2
t^2 = 29sec
t = 5.39sec

Add this to the time-to-apex to get answer D: 8.39sec

C is an extension of v=at
v1 = at + v0 and in this problem a is negative, since it's going against gravity initially.
v1 = v0 - at
v1 = 29.4 - 9.8*2
v1 = 29.4 - 19.6
v1 = 9.8 meters per second.

(interesting number - you could logically deduce this, given that the last one second of acceleration would be 1 second's worth of G, which would be 9.8m/sec)

distance for 2 seconds would be 1 second's worth of g from apex... s=1/2at^2 = 0.5 * 9.8 * 1 = 9.8/2 = 4.9m = 137.2m

C: 137.2m distance and speed = 9.8m/sec

EDIT: subtract the 98m height from answer C, since the question asks for distance, not height. Distance = 39.2

Answer 3

a. 2.99s
b. 142.26m
c. 39.18m @ 10.19m/s upward
d. 8.44s